Answer:
Given that
V2/V1= 0.25
And we know that in adiabatic process
TV^န-1= constant
So
T1/T2=( V1 /V2)^ န-1
So = ( 1/0.25)^ 0.66= 2.5
Also PV^န= constant
So P1/P2= (V2/V1)^န
= (1/0.25)^1.66 = 9.98
A. RMS speed is
Vrms= √ 3RT/M
But this is also
Vrms 2/Vrms1= (√T2/T1)
Vrms2=√2.5= 1.6vrms1
B.
Lambda=V/4π√2πr²N
So
Lambda 2/lambda 1= V2/V1 = 0.25
So the mean free path can be inferred to be 0.25 times the first mean free path
C. Using
Eth= 3/2KT
So Eth2/Eth1= T2/T1
So
Eth2= 2.5Eth1
D.
Using CV= 3/2R
Cvf= Cvi
So molar specific heat constant does not change
Hi there!
We can begin by solving for the time it takes the projectile to reach the ground.
Use the equation:
Plug in the given height:
Use the equation:
d = vt, to solve for the horizontal distance:
d = (54)(8.536) = 460.944 m
Answer: The amplitude is 0. (assuming that the amplitude ot both initial waves is the same)
Explanation:
When two monochromatic light waves of the same wavelength and same amplitude undergo destructive interference, means that the peak of one of the waves coincides with the trough of the other, so the waves "cancel" each other in that point in space.
Then if two light waves undergo destructive interference, the amplitude of the resultant wave in that particular point is 0.
Answer:
W=Fd
This is the equation for it