Answer:
a) t = 0.74s
b) D = 4.76m
c) Vf = 5.35m/s
Explanation:
The ball starts rolling when Vf = ωf*R.
We know that:
Vf = Vo - a*t
ωf = ωo + α*t
With a sum of forces on the ball:




With a sum of torque on the ball:



Replacing both accelerations:


t=0.74s
The distance will be:


Final velocity:

Vf=5.35m/s
Given :
Initial velocity, u = 12.5 m/s.
Height of camera, h = 64.3 m.
Acceleration due to gravity, g = 9.8 m/s².
To Find :
How long does it take the camera to reach the ground.
Solution :
By equation of motion :

Putting all given values, we get :

t = 2.56 and t = −5.116.
Since, time cannot be negative.
t = 2.56 s.
Therefore, time taken is 2.56 s.
Hence, this is the required solution.
Explanation:
It is given that,
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, 
Speed of light, 
Let t is the time interval required for the strong interaction to occur. The speed is given by :




So, the time interval required for the strong interaction to occur is
. Hence, this is the required solution.
Ocean bulges on Earth would be bigger if the Moon had twice as much mass and yet orbited the planet at the same distance. Option B is correct.
<h3>What is ocean bludge?</h3>
The fluid and moveable ocean water are drawn towards the moon by the gravitational attraction between the moon and the Earth.
The ocean nearest to the moon experiences a bulge as a result, and as the Earth rotates, the affected seas' locations shift.
The Moon's bulges in the oceans would be larger if it had twice the mass and orbited Earth at the same distance.
Hence option B is corect.
To learn more about the ocean bulge refer;
brainly.com/question/14373016
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