A flat surface of area 3.20m² is rotated in a uniform electric field of magnitude E=6.20 × 10⁵N . m²/C . Determine the electric
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Answer:
22 revolutions
Explanation:
2 rev/s = 2*(2π rad/rev) = 12.57 rad/s
The angular acceleration when it starting
The angular acceleration when it stopping:
The angular distance it covers when starting from rest:
The angular distance it covers when coming to complete stop:
So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions
The mechanical energy for the first and the second ball is
Mass of the first ball = 20 kg
The initial speed at which a cannonball is fired from a cannon =1000 m/s
The angle made by the cannonball while being fired from the cannon = 37°
The maximum height reached by the first ball is,
The maximum height of the first cannonball is 17478.69 m.
The initial speed at which a cannonball is fired from a cannon =1000 m/s
The angle made by the cannonball while being fired from the cannon = 90 °
For the first ball, total mechanical energy= Potential energy at maximum height + kinetic energy at the maximum height
So, the total mechanical energy is,
= mgh \: + \frac{1}{2}mv {}^{2} _{x}[/tex]
Therefore, the total mechanical energy for the first and the
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