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
I would assume air resistance is negligible and so the acceleration of the package would be approximately 9.81 m/s².
Taking downwards as positive, use v²=u²+2as.
v²=(-2)²+2(9.81)(14)
v=16.7 m/s
First, we calculate for the weight of the object by multiplying the given mass by the acceleration due to gravity which is equal to 9.8 m/s²
Weight = (14 kg)(9.8 m/s²)
Weight = 137.2 N
The component of the weight that is along the surface of the inclined plane is equal to this weight times the sine of the given angle.
Weight = (137.2 N)(sin 52°)
weight = 108.1 N
The meters per second
+1t a second / 2t
The resultant force on the positive charge is mathematically given as
X=40N
<h3>What is the magnitude of the electrostatic force on the negative charge?</h3>
Question Parameters:
Three-point charges, two positive and one negative, each having a magnitude of 20
Generally, the -ve charge is mathematically given as

Q+=X
Therefore

X=40N
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