uhh nevermind I got it
took a little reasoning but not too difficult
isothermal so delta-T=0 i.e. delta-U=0
Q + W'=U-->Q=-W'
W'=(delta-KE)
delta-S= Q/T
thanks
oh haha ok maybe that's the assumption to make because otherwise calculating the delta-U would be difficult or is the delta-U of the entire system=0 i.e. unchanged?
wait since
Q + W'=delta-U, where W'=work done by nonconservative/other forces i.e. the work done by the surroundings on the system
W'=(delta-KE) + (delta-PE)...in this case PEi=0 and PEf=0
thus W'=1/2mvf^2- 1/2mvi^2 for both cars
I know just intuivitely that when a car crashes the KE of...
S(total)=S(car) + S(car 2) + S(environment) where the values are changes in entropy
delta-S=delta-q/T, where delta-q=heat and T is a constant temperature?
how do I relate velocities and masses to heat?
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Raindrops of mass 0.99 mg fall vertically at a constant speed of 9.0 m/s, striking a horizontal skylight at the rate of 1000 drops/s and draining off. The window size is 17.0 cm x 25.0 cm. Assume the collisions of the drops with the window are completely inelastic...
yes some help on this would be nice...
using the ideal gas law, since nr=constant=PiVi/Ti=PfVf/Tf, thus
PfVf=a constant
F=PA
Fspring=kx
then -kx-P(outside)A=P(inside)A
where A=pi(r^2)
but now I am stuck...I understand that initially P(outside)A=P(inside)A and thus kx=F(spring)=0 but I am not...
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