<h2>QUESTION:- </h2>
➜what is kepler's law??
Kepler gave the three laws or theorems of motion of the orbitals bodies
This law state that the celestial bodies revolves around the stars in elliptical orbit and star as a single focus.
Example :- Earth revolves around the Sun as assuming it as single focus
This also shows that earth revolves around the sun in elliptical orbit.
Area covered by the planet is equal in equal duration of time irrespective of the position of the planet.
It also states that Angular momentum is constant
As Angular momentum is constant it means areal velocity is also constant.
where:-
A is the area.
T is the time.
L is the angular momentum.
M is the mass of the body.
square of the time of the revolution is directly proportional to the cube of the distance between the planet and star in Astronomical unit.
where:-
T = time of revolution
a is the distance between the planet and star.
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 answer is 21m because the motion is in one dimension with constant acceleration.
The initial velocity is 0, because it started from rest, the acceleration <span>ax</span> is <span>4.7<span>m<span>s2</span></span></span>, and the time t is <span>3.0s</span>
Plugging in our known values, we have
<span>Δx=<span>(0)</span><span>(3.0s)</span>+<span>12</span><span>(4.7<span>m<span>s2</span></span>)</span><span><span>(3.0s)</span>2</span>=<span>21<span>m</span></span></span>
