Answer:
Impulse, J = 9.34 kg-m/s
Explanation:
Given that,
Mass of the volleyball, m = 0.35 kg
Initial velocity, u = +2.09 m/s
Final speed, v = -24.6 m/s
We need to find the magnitude of the impulse that the player applies to the ball. We know that impulse is equal to the change in momentum . So,
So, the magnitude of the impulse that the player applies to the ball is 9.34 kg-m/s.
Answer:
1.44 x 10⁻⁶ C
Explanation:
= charge on one sphere
= charge on other sphere
= Total charge on the two spheres = 40 μC
=
= 40 x 10⁻⁶
= (40 x 10⁻⁶) -
eq-1
= distance between the two spheres = 50 cm = 0.50 m
= magnitude of force between the two spheres = 2.0 N
Magnitude of force between the two spheres is given as
= 1.44 x 10⁻⁶ C
Answer:
<em>The penny will hit the ground at 6.39 seconds</em>
Explanation:
<u>Free Fall</u>
The penny is dropped from a height of y=200 m. The equation of the height on a free-fall motion is given by:
Where , and t is the time.
Solving for t:
Using the value y=200:
t=6.39 sec
The penny will hit the ground at 6.39 seconds