Q = C.v
v = Q/C
v = 4 × 10^(-10)/250
= 4 × 10^(-10)/2.5 × 10^2
= 1.6 × 10^(-12) volt
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 study of sound is called sonics and the study of sound waves are acoustics
1. D - sound travels the fastest through solids
2. 50 mm/s - v=fa
3. B - only process that involves changing waves
Answer:
a.Beth
b.2232 s
Explanation:
We are given that
Distance,d=400 mi
Speed of Alan,v=45 mph
Speed of Beth,v'=55 mph
a.Time =
Using the formula
Time taken by Alan=
Time taken by Beth=
Alan will reach San Francisco at 4:53 PM
Beth will reach San Francisco at 4:16 PM
Beth will reach before Alan.
b.Difference between time=8.89-7.27=1.62 hr
t=1.62 hr
1.62-1=0.62 hr
0.62 hr=
Hence, Beth has to wait 2232 s for Alan to arrive .