<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" media="screen" href="/~d/styles/atom10full.xsl"?><?xml-stylesheet type="text/css" media="screen" href="http://feeds.feedburner.com/~d/styles/itemcontent.css"?><feed xmlns="http://www.w3.org/2005/Atom" xmlns:openSearch="http://a9.com/-/spec/opensearch/1.1/" xmlns:blogger="http://schemas.google.com/blogger/2008" xmlns:georss="http://www.georss.org/georss" xmlns:gd="http://schemas.google.com/g/2005" xmlns:thr="http://purl.org/syndication/thread/1.0" gd:etag="W/&quot;A0MMRHk-eyp7ImA9WhBbFE0.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758</id><updated>2013-05-13T02:44:45.753+01:00</updated><category term="statics" /><category term="dynamics" /><category term="announcement" /><category term="heat" /><category term="kinematics" /><category term="waves" /><category term="measurements" /><category term="QnA" /><category term="mathematics" /><category term="article" /><category term="NPCMY" /><category term="materials" /><category term="SHSPC" /><category term="tip" /><category term="problem" /><title>High School Physics Lab</title><subtitle type="html">Problems, tips and related material for high school physics students.</subtitle><link rel="http://schemas.google.com/g/2005#feed" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/posts/default" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/" /><author><name>InfoFries Webmaster</name><uri>http://www.blogger.com/profile/13676244902205078501</uri><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><generator version="7.00" uri="http://www.blogger.com">Blogger</generator><openSearch:totalResults>20</openSearch:totalResults><openSearch:startIndex>1</openSearch:startIndex><openSearch:itemsPerPage>25</openSearch:itemsPerPage><atom10:link xmlns:atom10="http://www.w3.org/2005/Atom" rel="self" type="application/atom+xml" href="http://feeds.feedburner.com/HighSchoolPhysicsLab" /><feedburner:info xmlns:feedburner="http://rssnamespace.org/feedburner/ext/1.0" uri="highschoolphysicslab" /><atom10:link xmlns:atom10="http://www.w3.org/2005/Atom" rel="hub" href="http://pubsubhubbub.appspot.com/" /><entry gd:etag="W/&quot;CkIASX8_fip7ImA9WxZQEEw.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-7721413102680656995</id><published>2008-02-13T14:56:00.017Z</published><updated>2008-02-14T17:35:48.146Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2008-02-14T17:35:48.146Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="QnA" /><category scheme="http://www.blogger.com/atom/ns#" term="materials" /><title>Q&amp;A: Finding Volume of a Floating Object</title><summary>The following question was asked in the comments section of the article Archimedes' Principle Explained.Q: Can you just explain how to measure the volume of an irregularly shaped object which floats in water? Can we use Archimedes' principle? How?A: It is certainly possible to do so using Archimedes' principle, provided the density of the object is known. The key here is to find the mass of the </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/7721413102680656995/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=7721413102680656995" title="1 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/7721413102680656995?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/7721413102680656995?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2008/02/q-finding-volume-of-floating-object.html" title="Q&amp;A: Finding Volume of a Floating Object" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><thr:total>1</thr:total></entry><entry gd:etag="W/&quot;CUAEQXw5fip7ImA9WB9VFU8.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-8314279654788329338</id><published>2007-12-01T15:30:00.000Z</published><updated>2007-12-01T15:41:40.226Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2007-12-01T15:41:40.226Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="NPCMY" /><category scheme="http://www.blogger.com/atom/ns#" term="materials" /><title>Problem: Materials #4</title><summary>A cube of ice is floating in a salt solution of density 1.25 times the density of pure water in a beaker. When the ice melts, the level of the solution in the vesselA. risesB. fallsC. remains unchangedD. falls at first and then rises to the same height as beforeSolutionShow solution &gt;&gt;The answer is A.A non-mathematical explanation would be that since the ice is floating, its mass would be the </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/8314279654788329338/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=8314279654788329338" title="2 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/8314279654788329338?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/8314279654788329338?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/12/problem-materials-4.html" title="Problem: Materials #4" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_ans_q12.gif" height="72" width="72" /><thr:total>2</thr:total></entry><entry gd:etag="W/&quot;DEICR3w7cCp7ImA9WB9QEU4.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-7358273538290712225</id><published>2007-10-21T06:31:00.000+01:00</published><updated>2007-10-23T11:49:26.208+01:00</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2007-10-23T11:49:26.208+01:00</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="NPCMY" /><category scheme="http://www.blogger.com/atom/ns#" term="materials" /><title>Problem: Materials #3</title><summary>Within a certain type of star called a neutron star, the  material at the centre has a mass density of 1.0 × 1018 kg m-3.  If a small sphere of this material of radius 1.0 × 10-5 m were  somehow transported to the surface of the earth, what would be the weight of  this sphere? (Assume that g = 10 m s-2)A. 1000 NB. 4200 NC. 4.2 × 104 ND. 7.0 × 104 NE. 3.1 × 109 NSolutionShow solution &gt;&gt;The answer </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/7358273538290712225/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=7358273538290712225" title="1 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/7358273538290712225?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/7358273538290712225?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/10/problem-materials-3.html" title="Problem: Materials #3" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_sphere_volume.gif" height="72" width="72" /><thr:total>1</thr:total></entry><entry gd:etag="W/&quot;DEAEQXk7fyp7ImA9WB9QEU4.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-6765071776987646011</id><published>2007-09-29T18:11:00.000+01:00</published><updated>2007-10-23T11:51:40.707+01:00</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2007-10-23T11:51:40.707+01:00</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="NPCMY" /><category scheme="http://www.blogger.com/atom/ns#" term="kinematics" /><title>Problem: Kinematics #4</title><summary>A train covers half the distance of its journey with a speed 20 m s-1 and the other half with a speed of 40 m s-1. The average speed of the train during the whole journey isA. 25 m s-1B. 27 m s-1C. 30 m s-1D. 32 m s-1E. 35 m s-1SolutionShow solution &gt;&gt;The answer is B.Let the total distance travelled by the train be d and the time taken for it to complete the first half and second half of the </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/6765071776987646011/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=6765071776987646011" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/6765071776987646011?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/6765071776987646011?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/09/problem-kinematics-4.html" title="Problem: Kinematics #4" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_ans_q10_1.gif" height="72" width="72" /><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;CkIBQHg_fyp7ImA9WB5VGE4.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-1860169189552499640</id><published>2007-08-11T11:22:00.000+01:00</published><updated>2007-08-11T12:09:11.647+01:00</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2007-08-11T12:09:11.647+01:00</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="statics" /><category scheme="http://www.blogger.com/atom/ns#" term="article" /><category scheme="http://www.blogger.com/atom/ns#" term="dynamics" /><title>The Concept of… Force</title><summary>Force is one of the most fundamental concepts of physics. The concept of force has formed part of statics and dynamics since ancient times. The modern concept of force is commonly explained in terms of Newton's three laws of motion set forth in his Principia Mathematica (1687).What is force?Force can be described as a push or pull. A force upon an object is always a result of interaction with </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/1860169189552499640/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=1860169189552499640" title="7 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/1860169189552499640?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/1860169189552499640?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/08/concept-of-force.html" title="The Concept of… Force" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_billard_collision.jpg" height="72" width="72" /><thr:total>7</thr:total></entry><entry gd:etag="W/&quot;C0cDSX4_eCp7ImA9WB5WGEU.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-2724060623298609809</id><published>2007-07-15T06:33:00.000+01:00</published><updated>2007-07-31T12:24:38.040+01:00</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2007-07-31T12:24:38.040+01:00</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="SHSPC" /><category scheme="http://www.blogger.com/atom/ns#" term="heat" /><title>Problem: Heat #1</title><summary>The air in a room consists principally of nitrogen and oxygen. The molecular weight of nitrogen (N2) is 28 g/mole and the molecular weight of oxygen (O2) is 32 g/mole. The air is in thermal equilibrium. Which statement is true?A.The average speeds of the nitrogen and oxygen molecules are equal.B.The average kinetic energies of the nitrogen and oxygen molecules are equal.C.The oxygen molecules </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/2724060623298609809/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=2724060623298609809" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/2724060623298609809?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/2724060623298609809?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/07/problem-heat-1.html" title="Problem: Heat #1" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;CU8MSH07fCp7ImA9WB9VFU8.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-5914019804752659695</id><published>2007-06-30T12:24:00.000+01:00</published><updated>2007-12-01T15:44:49.304Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2007-12-01T15:44:49.304Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="NPCMY" /><category scheme="http://www.blogger.com/atom/ns#" term="dynamics" /><title>Problem: Dynamics #1</title><summary>A stone thrown vertically upwards from the Earth’s surface travels until a height of h and falls back to its original position. If air resistance can be ignored, which graph below best represents the change of force, F which acts on the stone with its distance travelled, s?A.C.B.D.SolutionShow solution &gt;&gt; The answer is D. Because the stone is thrown upwards, the gravitational field strength of </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/5914019804752659695/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=5914019804752659695" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/5914019804752659695?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/5914019804752659695?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/06/problem-dynamics-1.html" title="Problem: Dynamics #1" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_figure_q8_a.gif" height="72" width="72" /><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;D0UEQHozeip7ImA9WxZRGUw.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-5165824189077723703</id><published>2007-06-02T16:58:00.001+01:00</published><updated>2008-02-13T15:06:41.482Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2008-02-13T15:06:41.482Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="NPCMY" /><category scheme="http://www.blogger.com/atom/ns#" term="measurements" /><title>Problem: Measurements #1</title><summary>x1,  y1  and x2,  y2 are two sets of values that  are related according to the following equation:The points (x1,  y1) and  (x2,  y2) lie on the graphA.  C.  B.  D. SolutionShow solution &gt;&gt;The answer is C.The above equation can be rearranged in this way:When xy = k, that means y is inversely proportional to x.  The graph y versus x will be like the one in C. Hide solution &lt;&lt;Question source: </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/5165824189077723703/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=5165824189077723703" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/5165824189077723703?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/5165824189077723703?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/06/problem-measurements-1.html" title="Problem: Measurements #1" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_eq_q7.gif" height="72" width="72" /><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;D0YDQng8fSp7ImA9WxZRGUw.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-5288486387001283964</id><published>2007-05-29T17:27:00.001+01:00</published><updated>2008-02-13T15:06:13.675Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2008-02-13T15:06:13.675Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="mathematics" /><category scheme="http://www.blogger.com/atom/ns#" term="article" /><category scheme="http://www.blogger.com/atom/ns#" term="measurements" /><title>The Relationship between Variables (Part II)</title><summary>This article is a continuation from Part I, which discusses about direct proportion. Inverse ProportionTwo sets of readings for the quantities x and y are given in the table below.x34612y4321Here, when x is doubled, y is halved; when x is trebled, y is one-third its previous value and so on.We say that y is inversely proportional to x, or y varies inversely as x, that is  The graph of y versus x </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/5288486387001283964/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=5288486387001283964" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/5288486387001283964?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/5288486387001283964?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/05/relationship-between-variables-part-ii.html" title="The Relationship between Variables (Part II)" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_y_ip_x.gif" height="72" width="72" /><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;DUcNQ3c9fyp7ImA9WB9TGUo.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-7669408726462116556</id><published>2007-05-28T05:57:00.000+01:00</published><updated>2007-09-28T11:58:12.967+01:00</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2007-09-28T11:58:12.967+01:00</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="NPCMY" /><category scheme="http://www.blogger.com/atom/ns#" term="kinematics" /><title>Problem: Kinematics #3</title><summary>The figure above shows a displacement versus time squared (t2) graph for the motion of an object.Which of the following motion can be represented by this graph?A. A ball that is travelling at terminal velocity.B. A ball that is falling freely from a stationary position.C. A ball that bounces back from the floor.D. A ball that is travelling on a rough surface.SolutionShow solution &gt;&gt; The answer is</summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/7669408726462116556/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=7669408726462116556" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/7669408726462116556?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/7669408726462116556?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/05/problem-kinematics-3.html" title="Problem: Kinematics #3" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_figure_q6.gif" height="72" width="72" /><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;D0YAQX04cCp7ImA9WxZRGUw.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-6420547762565829788</id><published>2007-05-26T06:09:00.002+01:00</published><updated>2008-02-13T15:05:40.338Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2008-02-13T15:05:40.338Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="mathematics" /><category scheme="http://www.blogger.com/atom/ns#" term="article" /><category scheme="http://www.blogger.com/atom/ns#" term="measurements" /><title>The Relationship between Variables (Part I)</title><summary>One of the most important mathematical operations in physics is finding the relationship between variables. Through the study of these relationships, we can know how a change in one variable affects another variable, thus enabling us to make predictions and conclusions easily.For example, an automobile maker will want to know how the mass of a car affects its acceleration, so that they can design</summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/6420547762565829788/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=6420547762565829788" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/6420547762565829788?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/6420547762565829788?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/05/relationship-between-variables-part-i.html" title="The Relationship between Variables (Part I)" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_dp_yx.gif" height="72" width="72" /><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;DUYESHc7cCp7ImA9WB9TGUo.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-6610839885311000610</id><published>2007-05-25T05:46:00.000+01:00</published><updated>2007-09-28T11:58:29.908+01:00</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2007-09-28T11:58:29.908+01:00</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="NPCMY" /><category scheme="http://www.blogger.com/atom/ns#" term="kinematics" /><title>Problem: Kinematics #2</title><summary>The figure above shows two objects, P and Q moving with velocities 30 m s-1 and 20 m s-1 respectively towards each other on a straight line.How long, after that instant, will P and Q meet?A. 100.0 sB. 83.3 sC. 20.0 sD. 13.3 sSolutionShow solution &gt;&gt; The answer is C.Let the distance travelled by P and Q before they meet each other be d1 and d2 respectively. Here, we can form an equation:d1 + d2= </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/6610839885311000610/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=6610839885311000610" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/6610839885311000610?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/6610839885311000610?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/05/problem-kinematics-2.html" title="Problem: Kinematics #2" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_figure_q5.gif" height="72" width="72" /><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;DUYHQXgycSp7ImA9WB9TGUo.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-6014397035539678846</id><published>2007-05-11T12:27:00.000+01:00</published><updated>2007-09-28T11:58:50.699+01:00</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2007-09-28T11:58:50.699+01:00</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="NPCMY" /><category scheme="http://www.blogger.com/atom/ns#" term="materials" /><title>Problem: Materials #2</title><summary>The density of an object is ¾ of the density of water. If the object is floating on the water surface, then the ratio between the volume of object above the water surface and the volume of object below the water surface isA. 1:4B. 1:3C. 3:4D. 4:3SolutionShow solution &gt;&gt; The answer is B.This is a problem that requires the application of Archimedes’ principle.We know that the density of the object </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/6014397035539678846/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=6014397035539678846" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/6014397035539678846?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/6014397035539678846?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/05/problem-materials-2.html" title="Problem: Materials #2" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_ans_q4_1.gif" height="72" width="72" /><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;D0YERHg_eCp7ImA9WxZRGUw.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-2536025833280491040</id><published>2007-05-05T06:34:00.001+01:00</published><updated>2008-02-13T15:05:05.640Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2008-02-13T15:05:05.640Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="NPCMY" /><category scheme="http://www.blogger.com/atom/ns#" term="materials" /><title>Problem: Materials #1</title><summary>An object has a weight of 4 N in air, 3 N in water and 2.8 N in a salt solution. If the density of water is 1000 kg m-3, the density of the salt solution isA. 830 kg m-3B. 1070 kg m-3C. 1200 kg m-3D. 1430 kg m-3SolutionShow solution &gt;&gt; The answer is C.This is a problem that requires the application of Archimedes’ principle. The object has a weight of 4 N in air. Because the upthrust on the object</summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/2536025833280491040/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=2536025833280491040" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/2536025833280491040?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/2536025833280491040?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/05/problem-materials-1.html" title="Problem: Materials #1" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_ans_q3_1.gif" height="72" width="72" /><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;C0EMRH4_fyp7ImA9WxZRGE4.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-8502842187556046673</id><published>2007-04-30T16:50:00.000+01:00</published><updated>2008-02-12T15:54:45.047Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2008-02-12T15:54:45.047Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="article" /><category scheme="http://www.blogger.com/atom/ns#" term="materials" /><title>Archimedes' Principle Explained</title><summary>Archimedes' principle is a law that explains buoyancy or upthrust. It states thatWhen a body is completely or partially immersed in a fluid it experiences an upthrust, or an apparent loss in weight, which is equal to the weight of fluid displaced.According to a tale, Archimedes discovered this law while taking a bath. After making this discovery, he is said to have leapt out of his bathtub and </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/8502842187556046673/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=8502842187556046673" title="6 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/8502842187556046673?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/8502842187556046673?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/04/archimedes-principle-explained.html" title="Archimedes' Principle Explained" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_upthrust.png" height="72" width="72" /><thr:total>6</thr:total></entry><entry gd:etag="W/&quot;D0UHRXY_fip7ImA9WxZRGUw.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-2559463123669007807</id><published>2007-04-29T16:04:00.002+01:00</published><updated>2008-02-13T15:07:14.846Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2008-02-13T15:07:14.846Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="kinematics" /><title>Problem: Kinematics #1</title><summary>A stone is released from a height of 20 m and allowed to fall in a straight line towards the ground. Ignoring air resistance, calculatea) the time taken for the stone to reach the ground.b) the velocity of the stone just before it touches the ground(Assume that g = 10 m s-2)SolutionShow solution &gt;&gt;The answers for (a) and (b) are 2 seconds and 20 m s-1 respectively.This is a problem utilising the </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/2559463123669007807/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=2559463123669007807" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/2559463123669007807?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/2559463123669007807?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/04/problem-kinematics-1.html" title="Problem: Kinematics #1" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_ans_q2_a.gif" height="72" width="72" /><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;DkMCQH48fip7ImA9WxZRGUg.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-4264154972062603561</id><published>2007-04-28T06:58:00.002+01:00</published><updated>2008-02-14T02:01:01.076Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2008-02-14T02:01:01.076Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="article" /><category scheme="http://www.blogger.com/atom/ns#" term="kinematics" /><title>The Equations for Uniform Acceleration</title><summary>There are four equations for uniform acceleration (also known as the kinematic equations) which are used to describe the motion of an object:1.2.3.4.Where:u – initial velocityv – final velocitya – accelerationt – times – displacementYou may be wondering – why there are four equations instead of one? Notice that each equation has one variable missing. For example, the first equation doesn’t have </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/4264154972062603561/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=4264154972062603561" title="6 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/4264154972062603561?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/4264154972062603561?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/04/equations-for-uniform-acceleration.html" title="The Equations for Uniform Acceleration" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_vuat.gif" height="72" width="72" /><thr:total>6</thr:total></entry><entry gd:etag="W/&quot;D0UDRXw7cSp7ImA9WxZRGUw.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-596740573332294062</id><published>2007-04-22T07:01:00.001+01:00</published><updated>2008-02-13T15:07:54.209Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2008-02-13T15:07:54.209Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="problem" /><category scheme="http://www.blogger.com/atom/ns#" term="NPCMY" /><category scheme="http://www.blogger.com/atom/ns#" term="waves" /><title>Problem: Waves #1</title><summary>The frequency of oscillation of a simple pendulum is f1. When the length of the pendulum l is increased to 2l, the frequency of oscillation of the simple pendulum is f2. The value of the ratio f1/f2 isSolutionShow solution &gt;&gt; The answer is D.The formula for the periodic time, T of a simple pendulum is given byand we know thatTherefore, the formula for f of a simple pendulum is given byNow we know</summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/596740573332294062/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=596740573332294062" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/596740573332294062?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/596740573332294062?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/04/problem-waves-1.html" title="Problem: Waves #1" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="http://i167.photobucket.com/albums/u137/infofries/hspl/th_answersq1.gif" height="72" width="72" /><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;A0cESHo7cSp7ImA9WxZRGUw.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-7360407164190066991</id><published>2007-04-20T18:13:00.001+01:00</published><updated>2008-02-13T16:10:09.409Z</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2008-02-13T16:10:09.409Z</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="tip" /><title>Getting Started</title><summary>This post is about getting started in physics in general and also benefiting from the future contents of this blog as promised in the Introduction post.For high school students studying physics at school, you can use your textbook for reference should any need arises.For those who have none or intend to self-study, the following sites and online resources should provide you with texts on high </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/7360407164190066991/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=7360407164190066991" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/7360407164190066991?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/7360407164190066991?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/04/getting-started.html" title="Getting Started" /><author><name>W. Jaylee</name><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><thr:total>0</thr:total></entry><entry gd:etag="W/&quot;DUAAQng7cSp7ImA9WB5TEEU.&quot;"><id>tag:blogger.com,1999:blog-559180414868012758.post-6010491126579573428</id><published>2007-04-20T10:59:00.000+01:00</published><updated>2007-05-25T11:09:03.609+01:00</updated><app:edited xmlns:app="http://www.w3.org/2007/app">2007-05-25T11:09:03.609+01:00</app:edited><category scheme="http://www.blogger.com/atom/ns#" term="announcement" /><title>Introduction</title><summary>This is another blog brought to you by InfoFries. This blog aims to help high school physics students cope with their physics lessons, guide them through their examinations and perhaps lead them to success in the field of physics in the future. This shall be done by placing great emphasis on the understanding of basic concepts.The articles posted here are meant to supplement existing physics </summary><link rel="replies" type="application/atom+xml" href="http://hsphysicslab.blogspot.com/feeds/6010491126579573428/comments/default" title="Post Comments" /><link rel="replies" type="text/html" href="http://www.blogger.com/comment.g?blogID=559180414868012758&amp;postID=6010491126579573428" title="0 Comments" /><link rel="edit" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/6010491126579573428?v=2" /><link rel="self" type="application/atom+xml" href="http://www.blogger.com/feeds/559180414868012758/posts/default/6010491126579573428?v=2" /><link rel="alternate" type="text/html" href="http://hsphysicslab.blogspot.com/2007/04/introduction.html" title="Introduction" /><author><name>InfoFries Webmaster</name><uri>http://www.blogger.com/profile/13676244902205078501</uri><email>noreply@blogger.com</email><gd:image rel="http://schemas.google.com/g/2005#thumbnail" width="16" height="16" src="http://img2.blogblog.com/img/b16-rounded.gif" /></author><thr:total>0</thr:total></entry></feed>
