A stone is launched straight up by a slingshot. Its initial speed is 19.6 m/s, and the stone is 1.50 m above the ground when lau
1 answer:
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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:
Explanation:
We could use the following suvat equation:
where
s is the vertical displacement of the coin
v is its final velocity, when it hits the water
t is the time
g is the acceleration of gravity
Taking upward as positive direction, in this problem we have:
s = -1.2 m
And the coin reaches the water when
t = 1.3 s
Substituting these data, we can find v:
where the negative sign means the direction is downward.