I can't agree with the opinion that kinetic energy of a particle of gas does not correspond to temperature. Each degree of freedom has energy kT/2. Mean kinetic energy of molecules (regardless of number of atoms) means 3 degrees of freedom. It means that the mean kinetics energy of molecules is...
This interpretation is a bit missleading. Microscopic kinetic energy(of molecul movements) corresponds with temperature rather then with heat. Heat doesn't equal temperature. Heat is simply energy that a system gains or gives up. Strictly speaking, heat is transfer of energy. Thus heat is...
Hello everyone,
I've been wondering if I can calculate the apparent magnitude of The Moon since I know the apparent magnitude of The Sun(say -27mag) and reflection coefficient of Moon's surface(say 0.12).
We know the equation
\Delta m = -2.5 log_{10}(I/I_0)
Where I is the flux of the...
There is a chapter in Feynman Lectures on Physics called The Random Walk(41-4). I understand everything till the paragraph right after equation 41.18. I have no idea what he is trying to say. There is an equation 41.19, which is diff. eq. for object that is forced and is in a environment that...
Well
a = \sqrt{a_g} \\
x = \sqrt{\alpha}v
So I got
\frac {\arctan({\sqrt{\frac{\alpha}{a_g}} v})}{\sqrt{a_g}} = t
didn't I?
So
v = \frac {\tan(\sqrt{a_g}t)}{\sqrt{\frac{\alpha}{a_g}}}
IS IT RIGHT ?
Greetings everyone.
Can you help me please with solving this differential equation?
\large \frac{dv}{dt} = a_g + \alpha v^n
where \Large a_g \alpha n are constants. Artelnatively with specific n as n = 1, 2.
I have no idea what to do...
Thank you very much
Ok. So how would you calculate the deformation of a pad if you put an cylinder on it. How would you calculate the distribution of normal force along the curvature?
I understand that it's sth. like calculating efective elastic modulus of two serial springs.
\frac {1}{E} = \frac{1}{E_1} + \frac{1}{E_2}
E is elastic modulus.
It's because we can say it's like two springs. One is the object and second is the pad.
And it is almost all I know about it.
I...