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
Answer:
find V using d = M/V
Explanation:
F = d*V*g
d = density of fluid (in this case, 1000)
V = volume of object
g = gravity
Answer:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
Explanation:
Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:
(1)
Where:
, 
- Initial and final momemtums of the small object, measured in kilogram-meters per second.
, 
- Initial and final momentums of the big object, measured in kilogram-meters per second.
If we know that 
, 
and 
, then the final momentum of the big object is:
The magnitude of the large object's momentum change is:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
