Answer:
Suppose the micrometeoroid weighed 1 g = .001 kg
Suppose also the spacecraft were moving at 18,000 mph (1.5 hrs per rev)
Usually, the smaller particle would be moving but for simplicity suppose that it were stationary wrt the ground
v = 18000 miles / hr * 1500 m/mile / 3600 sec/hr = 7500 m/s
KE = 1/2 * .001 kg * (7500 m)^2 = 28,125 Joules
One can see that 28000 Joules could be damaging amount of energy
Answer:
0.3817 N
Explanation:
Remark
One thing is certain: the ball has a mass of 101 grams wherever it is in the universe. That is not true of the force. The force on the moon is a whole lot less than it is on earth, and maybe planet x as well.
Givens
m = 101 g
vi = 0 That's what at rest means.
t = 2.91 s
d = 16 m
F= ?
Formulas
d = vi*t + 1/2*a * t^2
Force = m * a
Solution
16 = 0 + 1/2 a * 2.91^2
16 = 4.234 a Divide by 4.234
16/4.234 = a
a = 3.779
F = m * a
a = 3.779
m = 101 g = 1 kg / 1000 grams
m = 0.101 kg
F = 0.101 * 3.779
F = 0.3817N
Can you give us the options…?
(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg
Fam
