Solution :
From the given data,
For the spherical shell is
where x is the radius of gyration and the acceleration of a rolling body on an inclined plane = a
Therefore,
= 0.6 x 9.81 x sin ( 29.7)
Δt = 1.411 s
Answer:
Approximately 0.0898 W/m².
Explanation:
The intensity of light measures the power that the light delivers per unit area.
The source in this question delivers a constant power of . If the source here is a point source, that of power will be spread out evenly over a spherical surface that is centered at the point source. In this case, the radius of the surface will be 9.6 meters.
The surface area of a sphere of radius is equal to . For the imaginary 9.6-meter sphere here, the surface area will be:
.
That power is spread out evenly over this 9.6-meter sphere. The power delivered per unit area will be:
.
Answer:
Magnification, m = -0.42
Explanation:
It is given that,
Height of diamond ring, h = 1.5 cm
Object distance, u = -20 cm
Radius of curvature of concave mirror, R = 30 cm
Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)
Using mirror's formula :
, f = focal length of the mirror
v = -8.57 cm
The magnification of a mirror is given by,
m = -0.42
So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.
The first two are always the reactants the products come after so they are last
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