Answer:
468449163762.0812 W
Explanation:
m = Mass = 
V = Volume =
r = Distance of sphere from isotropic point source of light = 0.5 m
R = Radius of sphere = 2 mm
= Density = 19 g/cm³
c = Speed of light = 
A = Area = 
I = Intensity = 
g = Acceleration due to gravity = 9.81 m/s²
Force due to radiation is given by

According to the question

The power required of the light source is 468449163762.0812 W
Answer:
e*P_s = 11 W
Explanation:
Given:
- e*P = 1.0 KW
- r_s = 9.5*r_e
- e is the efficiency of the panels
Find:
What power would the solar cell produce if the spacecraft were in orbit around Saturn
Solution:
- We use the relation between the intensity I and distance of light:
I_1 / I_2 = ( r_2 / r_1 ) ^2
- The intensity of sun light at Saturn's orbit can be expressed as:
I_s = I_e * ( r_e / r_s ) ^2
I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2
I_s = 11 W / e*a
- We know that P = I*a, hence we have:
P_s = I_s*a
P_s = 11 W / e
Hence, e*P_s = 11 W
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Answer:
So the acceleration of the child will be 
Explanation:
We have given angular speed of the child 
Radius r = 4.65 m
Angular acceleration 
We know that linear velocity is given by 
We know that radial acceleration is given by 
Tangential acceleration is given by

So total acceleration will be 