An intriguing question: what would give you the worst skin burn for the same contact time, dry ice or liquid N2O?and why? ]]>

a) How much time will they save when doing cycles? Does this depend on agreed-on distance d?

b) Determine conditions under which both X and Y will get to B at the same time and calculate number of cycles needed to accomplish this.

I need help to understand the concept and solution. ]]>

L = f*Nb*N1*N2/(4*pi*sigma_x*sigma_y)

(see eq. 16 in link: https://cds.cern.ch/record/941318/files/p361.pdf)

Where f=revolution frequency (I imagine this is the time it take for a bunch to go around the synchrotron/cyclotron/device), Nb=number of bunches per beam, N1 and N2=number of particles per bunch in beam 1 and beam 2 respectively, and sigma_x and sigma_y are the variances in x and y for the Gaussian.

The bunches are described by the distribution p(x,y,z)=C*exp(-x²/(2*sigma_x²)-y²/(2*sigma_y²)-z²/(2*sigma_z²)).

I calculate the overlap integral and get that the luminosity in one collision is:

L_collision = N1*N2/(8*pi*sigma_x*sigma_y)

I am almost certain that this is correct. However, to get the total luminosity, they then multiply this by f*Nb*2 !!! Could someone explain where this comes from?

In my opinion, if we were to multiply L_collision by the total number of collisions in the device, we should multiply it by f*2*Nb², or simply by Nb if we are only considering the collisions at a specific point in the device.

Thanks in advance. ]]>

Height Water - Range (cm) - Velocity (m/s)

11.8cm - 31 - 1.13

10.6cm - 29 - 1.05

8.6cm - 25 - 0.91

6.8cm - 22 - 0.8

5cm - 18 - 0.66

3.6cm - 14 - 0.51

2.7cm - 10 - 0.36 ]]>

If you emit white light at a "green" object, my understanding is that it will absorb the green light and reflect everything else. That is why if you shine a green light on a green object, it will appear as black, since little to no light is reflecting. However, we can interpret the emission of that green light as green, where the sole present wavelength is the green wavelength, but we still understand it as the same color. It may be as simple as "both are true" but I'm curious if there's something I'm missing or not. ]]>

I've got a problem involving a track and two cars. The problem appears to be a fairly simple simultaneous equation but actually turns out to be a bit more complex.

Two toy cars start from the same position and race around a track. Both cars accelerate for 3 seconds. Car A accelerates at a rate of 2m/s/s while car B accelerates at a rate of 3.2m/s/s. The radius of the track is 2.5m. At what time does car b overtake car A.

The problem in encountering in the maths is that car B completes a lap before overtaking and therefor its displacement resets to 0 while car A continues about its journey.

Anyone have any ideas? ]]>

Let me elaborate. Newtons first law states that an object at rest remains at rest and object in motion remains in motion at constant speed in a straight line unless acted upon by external force. Now observe the continuous motion in same speed could mean conservation of velocity.

Now third law. It says every action has a reaction. So here say a billiard ball hits another then the momentum is transferred and remains in conservation.

What are your views. Especially asked to studiot. ]]>

I need help calculating the flow coefficient through a pressure reducing valve DN20.

The values I currently have are:

Inlet Pressure: 5 bar

Outlet Pressure: 2.75 bar

Flow capacity: 322kg/hr

According to this website below, the density of steam at 5bar is 2.667 kg/m3

Saturated Steam

Using these values, I substituted it into this website to help me calculate the CV value and the result I got was 4.9.

https://www.tlv.com/global/AU/calcul...ce-cv-kvs.html

However, when I manually substitute all the data into the first equation of that page I get an answer of CV=5.6

Why is my flow coefficient so inaccurate? Thanks. ]]>

================================================== ===========================

Edit: I tried to download the paper ( a pdf) from the site and got a "dangerous website" warning from my anti-virus. I am PMing paradigm and you can ask him for the link directly.

This will work unless, of course, paradigm is a troll...

-Dan ]]>

a) (M1g/A1) +(M2g/A2)

b) (M1g/A1) +(M2g/A1)

c) (M1g/A2) + (M2g/A2)

d) (M1g/A1) - (M2g/A2)

There is no answer given. However, I'm bit confused why there is no option of ((M1+M2)*g)/(A1+A2). Since Fg = Fn and P = Fn/A, the normal force of the blocks would be (M1+M2)*g. That would be divided by the sum of the areas. How would I solve this? ]]>

Consider a pool reactor (Fig. 2-33) whose core is constructed from a number of vertical fuel plates of thickness 2L. Initially the system has the uniform temperature T; then assume that the constant nuclear internal energy u"' is uniformly generated in these plates. The heat transfer coefficient between the plates and the coolant is h. The temperature of the coolant remains constant, and the thickness of the plates is small compared with other dimensions. Thus, if the end effects are neglected, the heat transfer may be taken to be one-dimensional. We wish to formulate the unsteady temperature problem of the reactor.

Attachment 2260

The author starts with the lumped formulation using this control volume:

Attachment 2261

The lumped first law of thermodynamics applied to this system reduces to:

$\displaystyle \frac{dE}{dt}=-2Aq_n$

Then the author states:

$\displaystyle \frac{dE}{dt}=\rho(A.2L)c\frac{dT}{dt}-(A.2L)u'''$

u''' is the internal energy generation per unit volume.

Inserting the last equation into the first we get the lumped form:

$\displaystyle \rho c L \frac{dE}{dt}=-q_n+u'''L$

My question is: Where did this equation come from:

$\displaystyle \frac{dE}{dt}=\rho(A.2L)c\frac{dT}{dt}-(A.2L)u'''$

The attempt at a solution:

So we have $\displaystyle T_{\mu\nu}=\frac{\partial L}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}L$, where $\displaystyle \phi$ is the field that the Lagrangian depends on. If we do the given change on the Lagrangian, the change in $\displaystyle T_{\mu\nu}$ would be $\displaystyle \frac{\partial (\partial_\alpha X^\alpha)}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}\partial_\alpha X^\alpha =\partial_\alpha \frac{\partial X^\alpha}{\partial(\partial_\mu \phi)}\partial_\nu \phi-g_{\mu\nu}\partial_\alpha X^\alpha$. From here I thought of using this: $\displaystyle g_{\mu\nu}\partial_\alpha X^\alpha=g_{\mu\nu}\partial_\alpha \phi \frac{\partial X^\alpha}{\partial \phi}$ But I don't really know what to do from here. Mainly I don't know how to get rid of that $\displaystyle g_{\mu\nu}$. Can someone help me? ]]>