Answer:
A. We have that radius r = 4.00m intensity I = 8.00 W/m^
total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W
b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2
c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J
The mass would be 51 kg. So b is the right answer.
Hold on and let's discuss this realistically.
Because of gravity, there are two forces between the Earth and me. One draws me toward the Earth. The strength of that force is what I call my "weight". The other force draws the Earth toward me, and has the same strength.
The strength of these forces depends on the masses of the Earth and me. If the strength just tripled, that means that at least one of us just picked up a lot more mass. If the Earth suddenly became three times as massive, then the weight of everything and everybody on it would suddenly triple, and I'm pretty sure it would be the end of all of us before too long.
If it was only MY mass that suddenly tripled, that would mean that I had gone tearing through my house and the neighbour's house, eating everything in sight including the 2 couches, 3 dogs, and 6 TVs. Naturally, just as you would expect, my weight changed from 207 to 621, and my skin is stretched really tight.
ooohhh
r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
Answer:
189 m/s
Explanation:
The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.
So, F = W
mv²/r = mg
v² = gr
v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m
So, v = √gr
v = √(9.8 m/s² × 3.63 × 10³ m)
v = √(35.574 × 10³ m²/s²)
v = √(3.5574 × 10⁴ m²/s²)
v = 1.89 × 10² m/s
v = 189 m/s
