Imagine a 10kg block moving with a velocity of 20m/s to the left.
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when the ball hits the floor and bounces back the momentum of the ball changes.
the rate of change of momentum is the force exerted by the floor on it.
the equation for the force exerted is
f = rate of change of momentum
v is the final velocity which is - 3.85 m/s
u is initial velocity - 4.23 m/s
m = 0.622 kg
time is the impact time of the ball in contact with the floor - 0.0266 s
substituting the values
since the ball is going down, we take that as negative and ball going upwards as positive.
f = 189 N
the force exerted from the floor is 189 N
Answer:
Explanation:
Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:
We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges() and inversely proportional to the square of the distance(d) that separates them.
Replacing the given values, where k is the Coulomb constant: