It occurs as detrital grains in sedimentary rocks. It forms under extreme pressure.
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
The third one should do the trick
Answer:C
Explanation:
Given
weight of object is equal to weight of object
Suppose weight of Planck is W
suppose weight is at distance of x cm from wedge
balancing Torque
i.e. at a distance of 0.25L from the Left end
Since the two charts after the collision stick together, we are dealing with a perfectly inelastic collision.
First, we need to find the speed of the two charts after the collision. In order to do so, we consider the conservation of momentum:
m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v
We can solve for v, considering also that v₂=0
v = m₁·v₁ / (m₁ + m<span>₂)
= 0.31 </span>· 0.90 / (0.31 + 0.50)
= 0.34 m/s
The kinetic energy lost (which is transformed into bounding energy between the two charts) will be the difference between the total kinetic energy before the collision and after the collision:
ΔE = E₁ - E₂ = 1/2·m₁·v₁ - 1/2·(m₁ + m₂)·v
= 1/2(0.31)(0.90) - 1/2(0.81)(0.34)
= 0.1395 - 0.1377
= 0.0018J
Hence, the correct answer is ΔE = <span>0.0018J</span>