The temperature of the air above it
Answer:
1.28 x 10^4 N
Explanation:
m = 1500 kg, h = 450 km, radius of earth, R = 6400 km
Let the acceleration due to gravity at this height is g'
g' / g = {R / (R + h)}^2
g' / g = {6400 / (6850)}^2
g' = 8.55 m/s^2
The force between the spacecraft and teh earth is teh weight of teh spacecraft
W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N
Answer:
375 m.
Explanation:
From the question,
Work done by the frictional force = Kinetic energy of the object
F×d = 1/2m(v²-u²)..................... Equation 1
Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.
Make d the subject of the equation.
d = 1/2m(v²-u²)/F.................. Equation 2
Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.
Note: If is negative because it tends to oppose the motion of the object.
Substitute into equation 2
d = 1/2(60)(0²-25²)/-50
d = 30(-625)/-50
d = -18750/-50
d = 375 m.
Hence the it will slide before coming to rest = 375 m
