The velocity of the object s calculated as 22.1 m/s.
<h3>What is the speed of the object?</h3>
Given that we can write that;
v^2 = u^2 + 2gh
Now u = 0 m/s because the object was dropped from a height
v^2 = 2gh
v = √2 * 9.8 * 25
v = 22.1 m/s
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1. 0.16 N
The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

where
G is the gravitational constant
is the mass of the asteroid
m = 100 kg is the mass of the man
r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid
Substituting, we find

2. 1.7 m/s
In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

where v is the minimum speed required to stay in orbit.
Re-arranging the equation and solving for v, we find:

Answer:
mu = 0.56
Explanation:
The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v: final speed = 0m/s (the car stops)
v_o: initial speed in the interval of interest = 60km/h
= 60(1000m)/(3600s) = 16.66m/s
x: distance = 25m
BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

Furthermore, you use the relation between the friction force and the friction coefficient:

hence, the friction coefficient is 0.56