Physics

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2011-06-01T04:32:00.000-07:00

Johannes Kepler: The
Laws of Planetary Motion

In the interplay between quantitative observation and theoretical construction that characterizes the development of modern science, we have seen that Brahe was the master of the first but was deficient in the second. The next great development in the history of astronomy was the theoretical intuition of Johannes Kepler (1571-1630), a German who went to Prague to become Brahe's assistant.

Brahe's Data and Kepler

Kepler and Brahe did not get along well. Brahe apparently mistrusted Kepler, fearing that his bright young assistant might eclipse him as the premiere astonomer of his day. He therefore let Kepler see only part of his voluminous data.He set Kepler the task of understanding the orbit of the planet Mars, which was particularly troublesome. It is believed that part of the motivation for giving the Mars problem to Kepler was that it was difficult, and Brahe hoped it would occupy Kepler while Brahe worked on his theory of the Solar System. In a supreme irony, it was precisely the Martian data that allowed Kepler to formulate the correct laws of planetary motion, thus eventually achieving a place in the development of astronomy far surpassing that of Brahe.

Kepler and the Elliptical Orbits
Unlike Brahe, Kepler believed firmly in the Copernican system. In retrospect, the reason that the orbit of Mars was particularly difficult was that Copernicus had correctly placed the Sun at the center of the Solar System, but had erred in assuming the orbits of the planets to be circles. Thus, in the Copernican theory epicycles were still required to explain the details of planetary motion.It fell to Kepler to provide the final piece of the puzzle: after a long struggle, in which he tried mightily to avoid his eventual conclusion, Kepler was forced finally to the realization that the orbits of the planets were not the circles demanded by Aristotle and assumed implicitly by Copernicus, but were instead the "flattened circles" that geometers call ellipses (See adjacent figure; the planetary orbits are only slightly elliptical and are not as flattened as in this example.)
The irony noted above lies in the realization that the difficulties with the Martian orbit derive precisely from the fact that the orbit of Mars was the most elliptical of the planets for which Brahe had extensive data. Thus Brahe had unwittingly given Kepler the very part of his data that would allow Kepler to eventually formulate the correct theory of the Solar System and thereby to banish Brahe's own theory!

Some Properties of Ellipses
Since the orbits of the planets are ellipses, let us review a few basic properties of ellipses.1. For an ellipse there are two points called foci (singular: focus) such that the sum of the distances to the foci from any point on the ellipse is a constant. In terms of the diagram shown to the left, with "x" marking the location of the foci, we have the equation

a + b = constant
that defines the ellipse in terms of the distances a and b.2. The amount of "flattening" of the ellipse is termed the eccentricity. Thus, in the following figure the ellipses become more eccentric from left to right. A circle may be viewed as a special case of an ellipse with zero eccentricity, while as the ellipse becomes more flattened the eccentricity approaches one. Thus, all ellipses have eccentricities lying between zero and one.
The orbits of the planets are ellipses but the eccentricities are so small for most of the planets that they look circular at first glance. For most of the planets one must measure the geometry carefully to determine that they are not circles, but ellipses of small eccentricity. Pluto and Mercury are exceptions: their orbits are sufficiently eccentric that they can be seen by inspection to not be circles.3. The long axis of the ellipse is called the major axis, while the short axis is called the minor axis (adjacent figure). Half of the major axis is termed asemimajor axis. The length of a semimajor axis is often termed the size of the ellipse. It can be shown that the average separation of a planet from the Sun as it goes around its elliptical orbit is equal to the length of the semimajor axis. Thus, by the "radius" of a planet's orbit one usually means the length of the semimajor axis.

The Laws of Planetary Motion
Kepler obtained Brahe's data after his death despite the attempts by Brahe's family to keep the data from him in the hope of monetary gain. There is some evidence that Kepler obtained the data by less than legal means; it is fortunate for the development of modern astronomy that he was successful. Utilizing the voluminous and precise data of Brahe, Kepler was eventually able to build on the realization that the orbits of the planets were ellipses to formulate his Three Laws of Planetary Motion.
Kepler's First Law:

I. The orbits of the planets are ellipses, with the Sun at one focus of the ellipse.

Kepler's First Law is illustrated in the image shown above. The Sun is not at the center of the ellipse, but is instead at one focus (generally there is nothing at the other focus of the ellipse). The planet then follows the ellipse in its orbit, which means that the Earth-Sun distance is constantly changing as the planet goes around its orbit. For purpose of illustration we have shown the orbit as rather eccentric; remember that the actual orbits are much less eccentric than this.

Kepler's Second Law:

II. The line joining the planet to the Sun sweeps out equal areas in equal times as the planet travels around the ellipse.

Kepler's second law is illustrated in the preceding figure. The line joining the Sun and planet sweeps out equal areas in equal times, so the planet moves faster when it is nearer the Sun. Thus, a planet executes elliptical motion with constantly changing angular speed as it moves about its orbit. The point of nearest approach of the planet to the Sun is termed perihelion; the point of greatest separation is termed aphelion. Hence, by Kepler's second law, the planet moves fastest when it is near perihelion and slowest when it is near aphelion.

Kepler's Third Law:

III. The ratio of the squares of the revolutionary periods for two planets is equal to the ratio of the cubes of their semimajor axes:

In this equation P represents the period of revolution for a planet and R represents the length of its semimajor axis. The subscripts "1" and "2" distinguish quantities for planet 1 and 2 respectively. The periods for the two planets are assumed to be in the same time units and the lengths of the semimajor axes for the two planets are assumed to be in the same distance units.
Kepler's Third Law implies that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit. Thus, we find that Mercury, the innermost planet, takes only 88 days to orbit the Sun but the outermost planet (Pluto) requires 248 years to do the same.

Calculations Using Kepler's Third Law
A convenient unit of measurement for periods is in Earth years, and a convenient unit of measurement for distances is the average separation of the Earth from the Sun, which is termed an astronomical unit and is abbreviated as AU. If these units are used in Kepler's 3rd Law, the denominators in the preceding equation are numerically equal to unity and it may be written in the simple form
This equation may then be solved for the period P of the planet, given the length of the semimajor axis,

or for the length of the semimajor axis, given the period of the planet,
As an example of using Kepler's 3rd Law, let's calculate the "radius" of the orbit of Mars (that is, the length of the semimajor axis of the orbit) from the orbital period. The time for Mars to orbit the Sun is observed to be 1.88 Earth years. Thus, by Kepler's 3rd Law the length of the semimajor axis for the Martian orbit is
which is exactly the measured average distance of Mars from the Sun. As a second example, let us calculate the orbital period for Pluto, given that its observed average separation from the Sun is 39.44 astronomical units. From Kepler's 3rd Law
which is indeed the observed orbital period for the planet Pluto.

Universal Law of Gravitation

2011-06-01T04:31:00.000-07:00

Sir Isaac Newton: The
Universal Law of Gravitation

There is a popular story that Newton was sitting under an apple tree, an apple fell on his head, and he suddenly thought of the Universal Law of Gravitation. As in all such legends, this is almost certainly not true in its details, but the story contains elements of what actually happened.

What Really Happened with the Apple?
Probably the more correct version of the story is that Newton, upon observing an apple fall from a tree, began to think along the following lines: The apple is accelerated, since its velocity changes from zero as it is hanging on the tree and moves toward the ground. Thus, by Newton's 2nd Law there must be a force that acts on the apple to cause this acceleration. Let's call this force "gravity", and the associated acceleration the "accleration due to gravity". Then imagine the apple tree is twice as high. Again, we expect the apple to be accelerated toward the ground, so this suggests that this force that we call gravity reaches to the top of the tallest apple tree.
Sir Isaac's Most Excellent Idea
Now came Newton's truly brilliant insight: if the force of gravity reaches to the top of the highest tree, might it not reach even further; in particular, might it not reach all the way to the orbit of the Moon! Then, the orbit of the Moon about the Earth could be a consequence of the gravitational force, because the acceleration due to gravity could change the velocity of the Moon in just such a way that it followed an orbit around the earth.This can be illustrated with the thought experiment shown in the following figure. Suppose we fire a cannon horizontally from a high mountain; the projectile will eventually fall to earth, as indicated by the shortest trajectory in the figure, because of the gravitational force directed toward the center of the Earth and the associated acceleration. (Remember that an acceleration is a change in velocity and that velocity is a vector, so it has both a magnitude and a direction. Thus, an acceleration occurs if either or both the magnitude and the direction of the velocity change.)

But as we increase the muzzle velocity for our imaginary cannon, the projectile will travel further and further before returning to earth. Finally, Newton reasoned that if the cannon projected the cannon ball with exactly the right velocity, the projectile would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth, which is curving away at the same rate that the projectile falls. That is, the cannon ball would have been put into orbit around the Earth. Newton concluded that the orbit of the Moon was of exactly the same nature: the Moon continuously "fell" in its path around the Earth because of the acceleration due to gravity, thus producing its orbit.By such reasoning, Newton came to the conclusion that any two objects in the Universe exert gravitational attraction on each other, with the force having a universal form:

The constant of proportionality G is known as the universal gravitational constant. It is termed a "universal constant" because it is thought to be the same at all places and all times, and thus universally characterizes the intrinsic strength of the gravitational force.
The Center of Mass for a Binary System
If you think about it a moment, it may seem a little strange that in Kepler's Laws the Sun is fixed at a point in space and the planet revolves around it. Why is the Sun privileged? Kepler had rather mystical ideas about the Sun, endowing it with almost god-like qualities that justified its special place. However Newton, largely as a corollary of his 3rd Law, demonstrated that the situation actually was more symmetrical than Kepler imagined and that the Sun does not occupy a privileged postion; in the process he modified Kepler's 3rd Law.Consider the diagram shown to the right. We may define a point called the center of mass between two objects through the equations

where R is the total separation between the centers of the two objects. The center of mass is familiar to anyone who has ever played on a see-saw. The fulcrum point at which the see-saw will exactly balance two people sitting on either end is the center of mass for the two persons sitting on the see-saw.Here is a Center of Mass Calculator that will help you make and visualize calculations concerning the center of mass. (Caution: this applet is written under Java 1.1, which is only supported by the most recent browsers. It should work on Windows systems under Netscape 4.06 or the most recent version of Internet Explorer 4.0, but may not yet work on Mac or Unix systems or earlier Windows browsers.)

Newton's Modification of Kepler's Third Law
Because for every action there is an equal and opposite reaction, Newton realized that in the planet-Sun system the planet does not orbit around a stationary Sun. Instead, Newton proposed that both the planet and the Sun orbited around the common center of mass for the planet-Sun system. He then modified Kepler's 3rd Law to read,
where P is the planetary orbital period and the other quantities have the meanings described above, with the Sun as one mass and the planet as the other mass. (As in the earlier discussion of Kepler's 3rd Law, this form of the equation assumes that masses are measured in solar masses, times in Earth years, and distances in astronomical units.) Notice the symmetry of this equation: since the masses are added on the left side and the distances are added on the right side, it doesn't matter whether the Sun is labeled with 1 and the planet with 2, or vice-versa. One obtains the same result in either case.Now notice what happens in Newton's new equation if one of the masses (either 1 or 2; remember the symmetry) is very large compared with the other. In particular, suppose the Sun is labeled as mass 1, and its mass is much larger than the mass for any of the planets. Then the sum of the two masses is always approximately equal to the mass of the Sun, and if we take ratios of Kepler's 3rd Law for two different planets the masses cancel from the ratio and we are left with the original form of Kepler's 3rd Law:

Thus Kepler's 3rd Law is approximately valid because the Sun is much more massive than any of the planets and therefore Newton's correction is small. The data Kepler had access to were not good enough to show this small effect. However, detailed observations made after Kepler show that Newton's modified form of Kepler's 3rd Law is in better accord with the data than Kepler's original form.
Two Limiting Cases
We can gain further insight by considering the position of the center of mass in two limits. First consider the example just addressed, where one mass is much larger than the other. Then, we see that the center of mass for the system essentially concides with the center of the massive object:
This is the situation in the Solar System: the Sun is so massive compared with any of the planets that the center of mass for a Sun-planet pair is always very near the center of the Sun. Thus, for all practical purposes the Sun IS almost (but not quite) motionless at the center of mass for the system, as Kepler originally thought.However, now consider the other limiting case where the two masses are equal to each other. Then it is easy to see that the center of mass lies equidistant from the two masses and if they are gravitationally bound to each other, each mass orbits the common center of mass for the system lying midway between them:

This situation occurs commonly with binary stars (two stars bound gravitationally to each other so that they revolve around their common center of mass). In many binary star systems the masses of the two stars are similar and Newton's correction to Kepler's 3rd Law is very large.Implements Newton's modified form of Kepler's 3rd law for two objects (planets or stars) revolving around their common center of mass. By making one mass much larger than the other in this interactive animation you can illustrate the ideas discussed above and recover Kepler's original form of his 3rd Law where a less massive object appears to revolve around a massive object fixed at one focus of an ellipse.
These limiting cases for the location of the center of mass are perhaps familiar from our afore-mentioned playground experience. If persons of equal weight are on a see-saw, the fulcrum must be placed in the middle to balance, but if one person weighs much more than the other person, the fulcrum must be placed close to the heavier person to achieve balance.

Weight and the Gravitational Force
We have seen that in the Universal Law of Gravitation the crucial quantity is mass. In popular language mass and weight are often used to mean the same thing; in reality they are related but quite different things. What we commonly call weight is really just the gravitational force exerted on an object of a certain mass. We can illustrate by choosing the Earth as one of the two masses in the previous illustration of the Law of Gravitation:
Thus, the weight of an object of mass m at the surface of the Earth is obtained by multiplying the mass m by the acceleration due to gravity, g, at the surface of the Earth. The acceleration due to gravity is approximately the product of the universal gravitational constant G and the mass of the Earth M, divided by the radius of the Earth, r, squared. (We assume the Earth to be spherical and neglect the radius of the object relative to the radius of the Earth in this discussion.) The measured gravitational acceleration at the Earth's surface is found to be about 980 cm/second/second.
Mass and Weight
Mass is a measure of how much material is in an object, but weight is a measure of the gravitational force exerted on that material in a gravitational field; thus, mass and weight are proportional to each other, with the acceleration due to gravity as the proportionality constant. It follows that mass is constant for an object (actually this is not quite true, but we will save that surprise for our later discussion of the Relativity Theory), but weight depends on the location of the object. For example, if we transported the preceding object of mass m to the surface of the Moon, the gravitational acceleration would change because the radius and mass of the Moon both differ from those of the Earth. Thus, our object has mass m both on the surface of the Earth and on the surface of the Moon, but it willweigh much less on the surface of the Moon because the gravitational acceleration there is a factor of 6 less than at the surface of the Earth.

Animation1

2011-05-20T03:50:00.000-07:00

Average vs. Instantaneous Speed

During a typical trip to school, your car will undergo a series of changes in its speed. If you were to inspect the speedometer readings at regular intervals, you would notice that it changes often. The speedometer of a car reveals information about the instantaneous speed of your car. It shows your speed at a particular instant in time.

The instantaneous speed of an object is not to be confused with the average speed.Average speed is a measure of the distance traveled in a given period of time; it is sometimes referred to as the distance per time ratio. Suppose that during your trip to school, you traveled a distance of 5 miles and the trip lasted 0.2 hours (12 minutes). The average speed of your car could be determined as

On the average, your car was moving with a speed of 25 miles per hour. During your trip, there may have been times that you were stopped and other times that your speedometer was reading 50 miles per hour. Yet, on average, you were moving with a speed of 25 miles per hour.

Animation

2011-05-20T03:43:00.000-07:00

Direction of Acceleration and Velocity

Consider the motion of a Hot Wheels car down an incline, across a level and straight section of track, around a 180-degree curve, and finally along a final straight section of track. Such a motion is depicted in the animation below. The car gains speed while moving down the incline - that is, it accelerates. Along the straight sections of track, the car slows down slightly (due to air resistance forces). Again the car could be described as having an acceleration. Finally, along the 180-degree curve, the car is changing its direction; once more the car is said to have an acceleration due to the change in the direction. Accelerating objects have a changing velocity - either due to a speed change (speeding up or slowing down) or a direction change.

This simple animation above depicts some additional information about the car's motion. The velocity and acceleration of the car are depicted by vector arrows. The direction of these arrows are representative of the direction of the velocity and acceleration vectors. Note that the velocity vector is always directed in the same direction which the car is moving. A car moving eastward would be described as having an eastward velocity. And a car moving westward would be described as having a westward velocity.

The direction of the acceleration vector is not so easily determined. As shown in the animation, an eastward heading car can have a westward directed acceleration vector. And a westward heading car can have an eastward directed acceleration vector. So how can the direction of the acceleration vector be determined? A simple rule of thumb for determining the direction of the acceleration is that an object which is slowing down will have an acceleration directed in the direction opposite of its motion. Applying this rule of thumbwould lead us to conclude that an eastward heading car can have a westward directed acceleration vector if the car is slowing down.

Be careful when discussing the direction of the acceleration of an object; slow down, apply some thought and use the rule of thumb.

Answer2

2011-05-20T03:42:00.001-07:00

Seymour has an average speed of

(95 yd) / (10 min) = 9.5 yd/min

and an average velocity of

(55 yd, left) / (10 min) = 5.5 yd/min, left

Answer1

2011-05-20T03:41:00.002-07:00

The skier has an average speed of

(420 m) / (3 min) = 140 m/min

and an average velocity of

(140 m, right) / (3 min) = 46.7 m/min, right

Speed and Velocity

2011-05-20T03:41:00.000-07:00

Speed and Velocity

Just as distance and displacement have distinctly different meanings (despite their similarities), so do speed and velocity. Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. A fast-moving object has a high speed and covers a relatively large distance in a short amount of time. A slow-moving object has a low speed and covers a relatively small amount of distance in a short amount of time. An object with no movement at all has a zero speed.

Velocity is a vector quantity that refers to "the rate at which an object changes its position." Imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. While this might result in a frenzy of activity, it would result in a zero velocity. Because the person always returns to the original position, the motion would never result in a change in position. Since velocity is defined as the rate at which the position changes, this motion results in zero velocity. If a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. Every step must go into moving that person further from where he or she started. For certain, the person should never change directions and begin to return to the starting position.

Velocity is a vector quantity. As such, velocity is direction aware. When evaluating the velocity of an object, one must keep track of direction. It would not be enough to say that an object has a velocity of 55 mi/hr. One must include direction information in order to fully describe the velocity of the object. For instance, you must describe an object's velocity as being 55 mi/hr, east. This is one of the essential differences between speed and velocity. Speed is a scalar quantity and does not keep track of direction; velocity is a vector quantity and is direction aware.

The task of describing the direction of the velocity vector is easy. The direction of the velocity vector is simply the same as the direction that an object is moving. It would not matter whether the object is speeding up or slowing down. If an object is moving rightwards, then its velocity is described as being rightwards. If an object is moving downwards, then its velocity is described as being downwards. So an airplane moving towards the west with a speed of 300 mi/hr has a velocity of 300 mi/hr, west. Note that speed has no direction (it is a scalar) and the velocity at any instant is simply the speed value with a direction.

As an object moves, it often undergoes changes in speed. For example, during an average trip to school, there are many changes in speed. Rather than the speed-o-meter maintaining a steady reading, the needle constantly moves up and down to reflect the stopping and starting and the accelerating and decelerating. One instant, the car may be moving at 50 mi/hr and another instant, it might be stopped (i.e., 0 mi/hr). Yet during the trip to school the person might average 32 mi/hr. The average speed during an entire motion can be thought of as the average of all speedometer readings. If the speedometer readings could be collected at 1-second intervals (or 0.1-second intervals or ... ) and then averaged together, the average speed could be determined. Now that would be a lot of work. And fortunately, there is a shortcut. Read on.

Calculating Average Speed and Average Velocity

The average speed during the course of a motion is often computed using the following formula:

In contrast, the average velocity is often computed using this formula

Let's begin implementing our understanding of these formulas with the following problem:

Q: While on vacation, Lisa Carr traveled a total distance of 440 miles. Her trip took 8 hours. What was her average speed?

To compute her average speed, we simply divide the distance of travel by the time of travel.

That was easy! Lisa Carr averaged a speed of 55 miles per hour. She may not have been traveling at a constant speed of 55 mi/hr. She undoubtedly, was stopped at some instant in time (perhaps for a bathroom break or for lunch) and she probably was going 65 mi/hr at other instants in time. Yet, she averaged a speed of 55 miles per hour. The above formula represents a shortcut method of determining the average speed of an object.

Average Speed versus Instantaneous Speed

Since a moving object often changes its speed during its motion, it is common to distinguish between the average speed and the instantaneous speed. The distinction is as follows.

Instantaneous Speed - the speed at any given instant in time.
Average Speed - the average of all instantaneous speeds; found simply by a distance/time ratio.

You might think of the instantaneous speed as the speed that the speedometer reads at any given instant in time and the average speed as the average of all the speedometer readings during the course of the trip. Since the task of averaging speedometer readings would be quite complicated (and maybe even dangerous), the average speed is more commonly calculated as the distance/time ratio.

Moving objects don't always travel with erratic and changing speeds. Occasionally, an object will move at a steady rate with a constant speed. That is, the object will cover the same distance every regular interval of time. For instance, a cross-country runner might be running with a constant speed of 6 m/s in a straight line for several minutes. If her speed is constant, then the distance traveled every second is the same. The runner would cover a distance of 6 meters every second. If we could measure her position (distance from an arbitrary starting point) each second, then we would note that the position would be changing by 6 meters each second. This would be in stark contrast to an object that is changing its speed. An object with a changing speed would be moving a different distance each second. The data tables below depict objects with constant and changing speed.

Now let's consider the motion of that physics teacher again. The physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. The entire motion lasted for 24 seconds. Determine the average speed and the average velocity.

The physics teacher walked a distance of 12 meters in 24 seconds; thus, her average speed was 0.50 m/s. However, since her displacement is 0 meters, her average velocity is 0 m/s. Remember that the displacement refers to the change in position and the velocity is based upon this position change. In this case of the teacher's motion, there is a position change of 0 meters and thus an average velocity of 0 m/s.

Here is another example similar to what was seen before in the discussion of distance and displacement. The diagram below shows the position of a cross-country skier at various times. At each of the indicated times, the skier turns around and reverses the direction of travel. In other words, the skier moves from A to B to C to D.

Use the diagram to determine the average speed and the average velocity of the skier during these three minutes. When finished, click the button to view the answer.

As a last example, consider a football coach pacing back and forth along the sidelines. The diagram below shows several of coach's positions at various times. At each marked position, the coach makes a "U-turn" and moves in the opposite direction. In other words, the coach moves from position A to B to C to D.

What is the coach's average speed and average velocity? When finished, click the button to view the answer.

In conclusion, speed and velocity are kinematic quantities that have distinctly different definitions. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance (a scalar quantity) per time ratio. Speed is ignorant of direction. On the other hand, velocity is a vector quantity; it is direction-aware. Velocity is the rate at which the position changes. The average velocity is the displacement or position change (a vector quantity) per time ratio.

Answer3

2011-05-20T03:24:00.003-07:00

The displacement of the cars is somewhere near 0 miles since they virtually finish where they started. Yet the successful cars have covered a distance of 500 miles.

Answer2

2011-05-20T03:24:00.001-07:00

The displacement of the runners is 0 miles. While they have covered a distance of 10 miles, they are not "out of place" or displaced. They finish where they started. Round-trip motions always have a displacement of 0.

Answer1

2011-05-20T03:23:00.002-07:00

The coach covers a distance of

(35 yds + 20 yds + 40 yds) = 95 yards

and has a displacement of 55 yards, left.

Distance and Displacement

2011-05-20T03:23:00.000-07:00

Distance and Displacement

Distance and displacement are two quantities that may seem to mean the same thing yet have distinctly different definitions and meanings.

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion.
Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

To test your understanding of this distinction, consider the motion depicted in the diagram below. A physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North.

Even though the physics teacher has walked a total distance of 12 meters, her displacement is 0 meters. During the course of her motion, she has "covered 12 meters of ground" (distance = 12 m). Yet when she is finished walking, she is not "out of place" - i.e., there is no displacement for her motion (displacement = 0 m). Displacement, being a vector quantity, must give attention to direction. The 4 meters east cancels the 4 meters west; and the 2 meters south cancels the 2 meters north. Vector quantities such as displacement are direction aware. Scalar quantities such as distance are ignorant of direction. In determining the overall distance traveled by the physics teachers, the various directions of motion can be ignored.

Now consider another example. The diagram below shows the position of a cross-country skier at various times. At each of the indicated times, the skier turns around and reverses the direction of travel. In other words, the skier moves from A to B to C to D.

Use the diagram to determine the resulting displacement and the distance traveled by the skier during these three minutes. Then click the button to see the answer.

As a final example, consider a football coach pacing back and forth along the sidelines. The diagram below shows several of coach's positions at various times. At each marked position, the coach makes a "U-turn" and moves in the opposite direction. In other words, the coach moves from position A to B to C to D.

What is the coach's resulting displacement and distance of travel? Click the button to see the answer.

To understand the distinction between distance and displacement, you must know the definitions. You must also know that a vector quantity such as displacement is direction-aware and a scalar quantity such as distance is ignorant of direction. When an object changes its direction of motion, displacement takes this direction change into account; heading the opposite direction effectively begins to cancel whatever displacement there once was.

Check Your Understanding

1. What is the displacement of the cross-country team if they begin at the school, run 10 miles and finish back at the school?

2. What is the distance and the displacement of the race car drivers in the Indy 500?

Answer5

2011-05-20T03:12:00.003-07:00

This is a scalar; there is no direction listed for it.

Answer4

2011-05-20T03:12:00.001-07:00

This is a scalar; there is no direction listed for it.

Answer3

2011-05-20T03:11:00.003-07:00

This is a scalar; there is no direction listed for it.

Answer2

2011-05-20T03:11:00.001-07:00

This is a vector; a direction is listed for it.

Answer1

2011-05-20T03:10:00.001-07:00

This is a vector; a direction is listed for it.

Answer

2011-05-20T03:09:00.001-07:00

This is a scalar; there is no direction listed for it.

Scalars and Vectors

2011-05-20T03:08:00.000-07:00

Scalars and Vectors

Physics is a mathematical science. The underlying concepts and principles have a mathematical basis. Throughout the course of our study of physics, we will encounter a variety of concepts that have a mathematical basis associated with them. While our emphasis will often be upon the conceptual nature of physics, we will give considerable and persistent attention to its mathematical aspect.

The motion of objects can be described by words. Even a person without a background in physics has a collection of words that can be used to describe moving objects. Words and phrases such as going fast, stopped, slowing down, speeding up, and turning provide a sufficient vocabulary for describing the motion of objects. In physics, we use these words and many more. We will be expanding upon this vocabulary list with words such asdistance, displacement, speed, velocity, and acceleration. As we will soon see, these words are associated with mathematical quantities that have strict definitions. The mathematical quantities that are used to describe the motion of objects can be divided into two categories. The quantity is either a vector or a scalar. These two categories can be distinguished from one another by their distinct definitions:

Scalars are quantities that are fully described by a magnitude (or numerical value) alone.
Vectors are quantities that are fully described by both a magnitude and a direction.

The remainder of this lesson will focus on several examples of vector and scalar quantities (distance, displacement, speed, velocity, and acceleration). As you proceed through the lesson, give careful attention to the vector and scalar nature of each quantity. As we proceed through other units at The Physics Classroom Tutorial and become introduced to new mathematical quantities, the discussion will often begin by identifying the new quantity as being either a vector or a scalar.

Check Your Understanding

1. To test your understanding of this distinction, consider the following quantities listed below. Categorize each quantity as being either a vector or a scalar. Click the button to see the answer.

Quantity	Category
a. 5 m
b. 30 m/sec, East
c. 5 mi., North
d. 20 degrees Celsius
e. 256 bytes

Describing Motion with Words

2011-05-20T03:06:00.000-07:00

Introduction to the Language of Kinematics

A typical physics course concerns itself with a variety of broad topics. One such topic is mechanics - the study of the motion of objects. The first six units of The Physics Classroom tutorial will involve an investigation into the physics of motion. As we focus on the language, principles, and laws that describe and explain the motion of objects, your efforts should center on internalizing the meaning of the information. Avoid memorizing the information; and avoid abstracting the information from the physical world that it describes and explains. Rather, contemplate the information, thinking about its meaning and its applications.

Kinematics is the science of describing the motion of objects using words, diagrams, numbers, graphs, and equations. Kinematics is a branch of mechanics. The goal of any study of kinematics is to develop sophisticated mental models that serve to describe (and ultimately, explain) the motion of real-world objects.

In this lesson, we will investigate the words used to describe the motion of objects. That is, we will focus on the language of kinematics. The hope is to gain a comfortable foundation with the language that is used throughout the study of mechanics. We will study such terms as scalars, vectors, distance, displacement, speed, velocity and acceleration. These words are used with regularity to describe the motion of objects. Your goal should be to become very familiar with their meaning.

2011-05-18T11:24:00.001-07:00

Using Equations as a Recipe for Algebraic Problem-Solving

As discussed in a previous part of Lesson 2, total system momentum is conserved for collisions between objects in an isolated system. The momentum lost by one object is equal to the momentum gained by another object. For collisions occurring in an isolated system, there are no exceptions to this law. This law becomes a powerful tool in physics because it allows for predictions of the before- and after-collision velocities (or mass) of an object. In this portion of Lesson 2, the law of momentum conservation will be used to make such predictions. The law of momentum conservation will be combined with the use of a "momentum table" and some algebra skills to solve problems involving collisions occurring in isolated systems.

Consider the following problem:

A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg person who is at rest on ice. The person catches the ball and subsequently slides with the ball across the ice. Determine the velocity of the person and the ball after the collision.

Such a motion can be considered as a collision between a person and a medicine ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the person to gain momentum. After the collision, the ball and the person travel with the same velocity (v) across the ice.

If it can be assumed that the affect of friction between the person and the ice is negligible, then the collision has occurred in an isolated system. Momentum should be conserved and the post-collision velocity (v) can be determined using a momentum table as shown below.

	Before Collision	After Collision
Person	0	(60 kg) • v
Medicine ball	(15 kg) • (20 km/hr) = 300 kg • km/hr	(15 kg) • v
Total	300 kg • km/hr	300

Observe in the table above that the known information about the mass and velocity of the two objects was used to determine the before-collision momenta of the individual objects and the total momentum of the system. Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. Finally, the expressions 60 kg • v and 15 kg • v were used for the after-collision momentum of the person and the medicine ball. To determine v (the velocity of both the objects after the collision), the sum of the individual momentum of the two objects can be set equal to the total system momentum. The following equation results:

60 • v + 15 • v = 300

75 • v = 300

v = 4 km/hr

Using algebra skills, it can be shown that v = 4 km/hr. Both the person and the medicine ball move across the ice with a velocity of 4 km/hr after the collision. (NOTE: The unit km/hr is the unit on the answer since the original velocity as stated in the question had units of km/hr.)

Now consider a similar problem involving momentum conservation.

A 0.150-kg baseball moving at a speed of 45.0 m/s crosses the plate and strikes the 0.250-kg catcher's mitt (originally at rest). The catcher's mitt immediately recoils backwards (at the same speed as the ball) before the catcher applies an external force to stop its momentum. If the catcher's hand is in a relaxed state at the time of the collision, it can be assumed that no net external force exists and the law of momentum conservation applies to the baseball-catcher's mitt collision. Determine the post-collision velocity of the mitt and ball.

Before the collision, the ball has momentum and the catcher's mitt does not. The collision causes the ball to lose momentum and the catcher's mitt to gain momentum. After the collision, the ball and the mitt move with the same velocity (v).

The collision between the ball and the catcher's mitt occurs in an isolated system, total system momentum is conserved. Thus, the total momentum before the collision (possessed solely by the baseball) equals the total momentum after the collision (shared by the baseball and the catcher's mitt). The table below depicts this principle of momentum conservation.

	Before Collision	After Collision
Ball	0.15 kg • 45 m/s = 6.75 kg•m/s	(0.15 kg) • v
Catcher's Mitt	0	(0.25 kg) • v
Total	6.75 kg•m/s	6.75 kg•m/s

Observe in the table above that the known information about the mass and velocity of baseball and the catcher's mitt was used to determine the before-collision momenta of the individual objects and the total momentum of the system. Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. Finally, the expression 0.15 • v and 0.25 • v are used for the after-collision momentum of the baseball and catcher's mitt. To determine v (the velocity of both objects after the collision), the sum of the individual momentum of the two objects is set equal to the total system momentum. The following equation results:

0.15 kg • v + 0.25 kg • v = 6.75 kg•m/s

0.40 kg • v = 6.75 kg•m/s

v = 16.9 m/s

Using algebra skills, it can be shown that v = 16.9 m/s. Both the baseball and the catcher's mitt move with a velocity of 16.9 m/s immediately after the collision and prior to the moment that the catcher begins to apply an external force.

The two collisions above are examples of inelastic collisions. Technically, an inelastic collision is a collision in which the kinetic energy of the system of objects is not conserved. In an inelastic collision, the kinetic energy of the colliding objects is transformed into other non-mechanical forms of energy such as heat energy and sound energy. The subject ofenergy will be treated in a later unit of The Physics Classroom. To simplify matters, we will consider any collisions in which the two colliding objects stick together and move with the same post-collision speed to be an extreme example of an inelastic collision.

Now we will consider the analysis of a collision in which the two objects do not stick together. In this collision, the two objects will bounce off each other. While this is not technically an elastic collision, it is more elastic than the previous collisions in which the two objects stick together.

A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision.

In this collision, the truck has a considerable amount of momentum before the collision and the car has no momentum (it is at rest). After the collision, the truck slows down (loses momentum) and the car speeds up (gains momentum).

The collision can be analyzed using a momentum table similar to the above situations.

	Before Collision	After Collision
Truck	3000 • 10 = 30 000	3000 • v
Car	0	1000 • 15 = 15 000
Total	30 000	30 000

Observe in the table above that the known information about the mass and velocity of the truck and car was used to determine the before-collision momenta of the individual objects and the total momentum of the system. Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. The after-collision velocity of the car is used (in conjunction with its mass) to determine its momentum after the collision. Finally, the expression 3000•v was used for the after-collision momentum of the truck (v is the velocity of the truck after the collision). To determine v (the velocity of the truck), the sum of the individual after-collision momentum of the two objects is set equal to the total momentum. The following equation results:

**3000*v + 15 000 = 30 000**

**3000*v = 15 000**

v = 5.0 m/s

Using algebra skills, it can be shown that v = 5.0 m/s. The truck's velocity immediately after the collision is 5.0 m/s. As predicted, the truck has lost momentum (slowed down) and the car has gained momentum.

The three problems above illustrate how the law of momentum conservation can be used to solve problems in which the after-collision velocity of an object is predicted based on mass-velocity information. There are additional practice problems (with accompanying solutions) later in this lesson that are worth the practice. However, be certain that you don't come to believe that physics is merely an applied mathematics course that is devoid of concepts. For certain, mathematics is applied in physics. However, physics is about concepts and the variety of means in which they are represented. Mathematical representations are just one of the many representations of physics concepts. Avoid merely treating these collision problems as mere mathematical exercises. Take the time to understand the concept of momentum conservation that provides the basis of their solution.

The next section of this lesson involves examples of problems that provide a real test of your conceptual understanding of momentum conservation in collisions. Before proceeding with the practice problems, be sure to try a few of the more conceptual questions that follow.

2011-05-18T11:20:00.001-07:00

The Cart and The Brick

Part B

Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum lost by one object equals the momentum gained by the other object.

The animation below portrays the collision between a 1.0-kg cart and a 2-kg dropped brick. It will be assumed that there are no net external forces acting upon the two objects involved in the collision. The only net force acting upon the two objects (the cart and the dropped brick) are internal forces - the force of friction between the cart and the dropped brick. The before- and after-collision velocities and momentum are shown in the data tables.

In the collision between the cart and the dropped brick, total system momentum is conserved. Before the collision, the momentum of the cart is 60 kg*cm/s and the momentum of the dropped brick is 0 kg*cm/s; the total system momentum is 60 kg*cm/s. After the collision, the momentum of the cart is 20.0 kg*cm/s and the momentum of the dropped brick is 40.0 kg*cm/s; the total system momentum is 60 kg*cm/s. The momentum of the cart-dropped brick system is conserved. The momentum lost by the cart (40 kg*cm/s) is gained by the dropped brick.

2011-05-18T11:19:00.003-07:00

Momentum Conservation Principle

One of the most powerful laws in physics is the law ofmomentum conservation. The law of momentum conservation can be stated as follows.

For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

The above statement tells us that the total momentum of a collection of objects (a system) isconserved - that is, the total amount of momentum is a constant or unchanging value. This law of momentum conservation will be the focus of the remainder of Lesson 2. To understand the basis of momentum conservation, let's begin with a short logical proof.

Consider a collision between two objects - object 1 and object 2. For such a collision, the forces acting between the two objects are equal in magnitude and opposite in direction (Newton's third law). This statement can be expressed in equation form as follows.

The forces act between the two objects for a given amount of time. In some cases, the time is long; in other cases the time is short. Regardless of how long the time is, it can be said that the time that the force acts upon object 1 is equal to the time that the force acts upon object 2. This is merely logical. Forces result from interactions (or contact) between two objects. If object 1 contacts object 2 for 0.050 seconds, then object 2 must be contacting object 1 for the same amount of time (0.050 seconds). As an equation, this can be stated as

Since the forces between the two objects are equal in magnitude and opposite in direction, and since the times for which these forces act are equal in magnitude, it follows that theimpulses experienced by the two objects are also equal in magnitude and opposite in direction. As an equation, this can be stated as

But the impulse experienced by an object is equal to the change in momentum of that object (the impulse-momentum change theorem). Thus, since each object experiences equal and opposite impulses, it follows logically that they must also experience equal and opposite momentum changes. As an equation, this can be stated as

The above equation is one statement of the law of momentum conservation. In a collision, the momentum change of object 1 is equal to and opposite of the momentum change of object 2. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. In most collisions between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object 1 loses 75 units of momentum, then object 2 gains 75 units of momentum. Yet, the total momentum of the two objects (object 1 plus object 2) is the same before the collision as it is after the collision. The total momentum of the system (the collection of two objects) is conserved.

A useful analogy for understanding momentum conservation involves a money transaction between two people. Let's refer to the two people as Jack and Jill. Suppose that we were to check the pockets of Jack and Jill before and after the money transaction in order to determine the amount of money that each possesses. Prior to the transaction, Jack possesses $100 and Jill possesses $100. The total amount of money of the two people before the transaction is $200. During the transaction, Jack pays Jill $50 for the given item being bought. There is a transfer of $50 from Jack's pocket to Jill's pocket. Jack has lost $50 and Jill has gained $50. The money lost by Jack is equal to the money gained by Jill. After the transaction, Jack now has $50 in his pocket and Jill has $150 in her pocket. Yet, the total amount of money of the two people after the transaction is $200. The total amount of money (Jack's money plus Jill's money) before the transaction is equal to the total amount of money after the transaction. It could be said that the total amount of money of the system (the collection of two people) is conserved. It is the same before as it is after the transaction.

A useful means of depicting the transfer and the conservation of money between Jack and Jill is by means of a table.

The table shows the amount of money possessed by the two individuals before and after the interaction. It also shows the total amount of money before and after the interaction. Note that the total amount of money ($200) is the same before and after the interaction - it is conserved. Finally, the table shows the change in the amount of money possessed by the two individuals. Note that the change in Jack's money account (-$50) is equal to and opposite of the change in Jill's money account (+$50).

For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision. A common physics lab involves the dropping of a brick upon a cart in motion.

The dropped brick is at rest and begins with zero momentum. The loaded cart (a cart with a brick on it) is in motion with considerable momentum. The actual momentum of the loaded cart can be determined using the velocity (often determined by a ticker tape analysis) and the mass. The total amount of momentum is the sum of the dropped brick's momentum (0 units) and the loaded cart's momentum. After the collision, the momenta of the two separate objects (dropped brick and loaded cart) can be determined from their measured mass and their velocity (often found from a ticker tape analysis). If momentum is conserved during the collision, then the sum of the dropped brick's and loaded cart's momentum after the collision should be the same as before the collision. The momentum lost by the loaded cart should equal (or approximately equal) the momentum gained by the dropped brick. Momentum data for the interaction between the dropped brick and the loaded cart could be depicted in a table similar to the money table above.

	Before Collision Momentum	After Collision Momentum	Change in Momentum
Dropped Brick	0 units	14 units	+14 units
Loaded Cart	45 units	31 units	-14 units
Total	45 units	45 units

Note that the loaded cart lost 14 units of momentum and the dropped brick gained 14 units of momentum. Note also that the total momentum of the system (45 units) was the same before the collision as it was after the collision.

Collisions commonly occur in contact sports (such as football) and racket and bat sports (such as baseball, golf, tennis, etc.). Consider a collision in football between a fullback and a linebacker during a goal-line stand. The fullback plunges across the goal line and collides in midair with the linebacker. The linebacker and fullback hold each other and travel together after the collision. The fullback possesses a momentum of 100 kg*m/s, East before the collision and the linebacker possesses a momentum of 120 kg*m/s, West before the collision. The total momentum of the system before the collision is 20 kg*m/s, West (review the section on adding vectors if necessary). Therefore, the total momentum of the system after the collision must also be 20 kg*m/s, West. The fullback and the linebacker move together as a single unit after the collision with a combined momentum of 20 kg*m/s. Momentum is conserved in the collision. A vector diagram can be used to represent this principle of momentum conservation; such a diagram uses an arrow to represent the magnitude and direction of the momentum vector for the individual objects before the collision and the combined momentum after the collision.

Now suppose that a medicine ball is thrown to a clown who is at rest upon the ice; the clown catches the medicine ball and glides together with the ball across the ice. The momentum of the medicine ball is 80 kg*m/s before the collision. The momentum of the clown is 0 m/s before the collision. The total momentum of the system before the collision is 80 kg*m/s. Therefore, the total momentum of the system after the collision must also be 80 kg*m/s. The clown and the medicine ball move together as a single unit after the collision with a combined momentum of 80 kg*m/s. Momentum is conserved in the collision.

Momentum is conserved for any interaction between two objects occurring in an isolated system. This conservation of momentum can be observed by a total system momentum analysis or by a momentum change analysis. Useful means of representing such analyses include a momentum table and a vector diagram. Later in Lesson 2, we will use the momentum conservation principle to solve problems in which the after-collision velocity of objects is predicted.

2011-05-18T11:17:00.001-07:00

Two Cars in 2-Dimensional Collision

Inelastic Collision

Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum change of the individual objects are equal in magnitude and opposite in direction.

Certain collisions are referred to as elastic collisions. Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.

The animation below portrays the inelastic collision between two 1000-kg cars. The before- and after-collision velocities and momentum are shown in the data tables.

In the collision between the two cars, total system momentum is conserved. Yet this might not be apparent without an understanding of the vector nature of momentum. Momentum, like all vector quantities, has both a magnitude (size) and a direction. When considering the total momentum of the system before the collision, the individual momentum of the two cars must be added as vectors. That is, 20 000 kg*m/s, East must be added to 10 000 kg*m/s, North. The sum of these two vectors is not 30 000 kg*m/s; this would only be the case if the two momentum vectors had the same direction. Instead, the sum of 20 000 kg*m/s, East and 10 000 kg*m/s, North is 22 361 kg*m/s at an angle of 26.6 North of East. Since the two momentum vectors are at right angles, their sum can be found using the Pythagorean theorem; the direction can be found using SOH CAH TOA (specifically, the tangent function). The value 22 361 kg*m/s is the total momentum of the system before the collision; and since momentum is conserved, it is also the total momentum of the system after the collision. Since the cars have equal mass, the total system momentum is shared equally by each individual car. In order to determine the momentum of either individual car, this total system momentum must be divided by two (approx. 11 200 kg*m/s). Once the momentum of the individual cars are known, the after-collision velocity is determined by simply dividing momentum by mass (v=p/m).

An analysis of the kinetic energy of the two objects reveals that the total system kinetic energy before the collision is 250 000 Joules (200 000 J for the eastbound car plus 50 000 J for the northbound car). After the collision, the total system kinetic energy is 125 000 Joules (62 500 J for each car). The total kinetic energy before the collision is not equal to the total kinetic energy after the collision. A large portion of the kinetic energy is converted to other forms of energy such as sound energy and thermal energy. A collision in which total system kinetic energy is not conserved is known as an inelastic collision.

The Unification of Magnetism with Einstein’s

2011-04-30T07:22:00.001-07:00

The Unification of Magnetism with Einstein’s E=mc²

(Mathematically proven on EinsteinElectricity.com)

A giant discovery ! After thousands of years man finally knows "exactly" what causes

The Force of Magnetism.

The only means of producing Light is via the combustion of Mass. The mathematical equation for Light is Mass divided by Temperature (m/AZ) which is equal to Electrical-Mass divided by Time (q/Z). Mass divided by Temperature is the definition of absolute cold. When Mass burns, Electrical-Mass escapes and immediately begins to move towards Mass that possesses Temperature, this is why Electrical-Mass from the sun is responsible for our Gravity.
Gravity is the pressure of Electrical-Mass that is right now pushing you down onto the earth. As this Electrical-Mass moves into and through the earth it creates the earth's Magnetic Field just like the magnetic field around any common magnet.
The varying Frequencies and Wavelengths of Electrical-Mass is also responsible for the Electromagnetic spectrum.

At earth's orbit radius (which is determined by Orbit Time) (which is in turn determined by planet Mass Density) there's an "Electrical-Mass Flow" that passes through all Mass and into and through the earth, namelyGravity. Now when you take a piece of metal and turn it into a Magnet (via electricity you simply change its lattice structure).

The atomic structure is rearranged so that the flow ofElectrical-Mass passes through the structure in only one direction or else more easily in that one direction.

This is why an object becomes Magnetized. Nothing more than a simple tunnel for that "Electrical-Mass-Flow" that is passing through it. All of the Electrical-Mass around that metal bar (like water at the bottom of the ocean) is wanting to enter the bar via the pressure of Gravity. Electrical-Mass upon all of its surfaces is forcing itself into the metal bar and due to the rearrangement of the lattice structure the flow now occurs more easily in one direction. The bar has now become a very simple Electrical-Mass-Flow gun.

This flow of Electrical-Mass through the metal bar is the very definition of

Magnetism.

The two pieces of Mass are nothing more than directional tunnels of Electrical-Mass flow. This flow is continually happening all around us and through us (keep in mind that all Mass is 99% empty space). And so Electrical-Mass enters all surfaces of the top two bars as it then flows through each of the bars easier in the one direction as the Electrical-Mass is then pushed "at" each other creating the "repelling force" of Magnetism.
As well the flow into the lower two bars then pushes away from each other creating the "attraction force" of Magnetism (a simple example of the Bernoulli effect).
Electrical-Mass Flow is ultimately responsible for the two metal objects being either magnetically repelled or attracted to each other. Light and Heat from the Energy of the sun (Light x Heat = Energy) (q/Z x A²Z³ = q(AZ)² = qc² = Energy) is indeed transformed into Magnetism via the Pressure of the flow of Electrical-Mass upon the earth.

Magnetism is the Force of the "current" of the flow of Electrical-Mass through the altered lattice
of a piece of mass caused by the pressure of Gravity.

Heated air does not rise because it is hotter, it rises because it has a greater (A) value. Any particle of Mass with a greater A value has a higher temperature becauseTemperature = AxZ. Because Electrical-Mass Density isequal to q/A²Z⁵ we can see the relationship between Density and Gravity and that the hot air is really being "pushed less" downward by Gravity due to the particle'slesser density. The Electrical-Mass (Gravity) moving downward through hotter air moves more easily through it and therefore does not push as hard upon it and therefore it rises due to the colder (more dense) air around it being pushed down more readily.

Physics

ree FTP and Download Manager 6.1.0.1

Brahe's Data and Kepler

Kepler and the Elliptical Orbits

Some Properties of Ellipses

The Laws of Planetary Motion

Kepler's First Law:

Kepler's Second Law:

Kepler's Third Law:

Calculations Using Kepler's Third Law

Universal Law of Gravitation

What Really Happened with the Apple?

Sir Isaac's Most Excellent Idea

The Center of Mass for a Binary System

Newton's Modification of Kepler's Third Law

Two Limiting Cases

Weight and the Gravitational Force

Mass and Weight

Animation1

Average vs. Instantaneous Speed

Animation

Direction of Acceleration and Velocity

Answer2

Answer1

Speed and Velocity

Speed and Velocity

Calculating Average Speed and Average Velocity

Q: While on vacation, Lisa Carr traveled a total distance of 440 miles. Her trip took 8 hours. What was her average speed?

Average Speed versus Instantaneous Speed

Answer3

Answer2

Answer1

Distance and Displacement

Distance and Displacement

Check Your Understanding

Answer5

Answer4

Answer3

Answer2

Answer1

Answer

Scalars and Vectors

Scalars and Vectors

Check Your Understanding

Describing Motion with Words

Introduction to the Language of Kinematics

Using Equations as a Recipe for Algebraic Problem-Solving

Before Collision

After Collision

Person

Medicine ball

Total

60 • v + 15 • v = 300

75 • v = 300

v = 4 km/hr

Before Collision

After Collision

Ball

Catcher's Mitt

Total

0.15 kg • v + 0.25 kg • v = 6.75 kg•m/s

0.40 kg • v = 6.75 kg•m/s

v = 16.9 m/s

Before Collision

After Collision

Truck

Car

Total

3000*v + 15 000 = 30 000

3000*v = 15 000

v = 5.0 m/s

The Cart and The Brick

Part B

Momentum Conservation Principle

Before

Collision

Momentum

After

Collision

Momentum

**3000*v + 15 000 = 30 000**

**3000*v = 15 000**