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2 answers:
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(a) +9.30 kg m/s
The impulse exerted on an object is equal to its change in momentum:
where
m is the mass of the object
is the change in velocity of the object, with
v = final velocity
u = initial velocity
For the volleyball in this problem:
m = 0.272 kg
u = -12.6 m/s
v = +21.6 m/s
So the impulse is
(b) 155 N
The impulse can also be rewritten as
where
F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)
is the duration of the collision
In this situation, we have
So we can re-arrange the equation to find the magnitude of the average force:
Answer:
The ball traveled 0.827 m
Explanation:
Given;
distance between the metal plates of the room, d = 3.1 m
mass of the glass, m = 1.1g
charge on the glass, q = 4.7 nC
speed of the glass ball, v = 4.8 m/s
voltage of the ceiling, V = +3.0 x 10⁶ V
The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;
F = qV/d
|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)
|F| = 4.548 x 10⁻³ N
F = - 4.548 x 10⁻³ N
The net horizontal force experienced by this ball is;
The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.
W = K.E
Therefore, the ball traveled 0.827 m