A 20 kg ball is thrown into the air. It is going 5.0 m/s when thrown. How much potential energy will it have at the top

1 answer:

5 0

When the ball leaves your hand, it has kinetic energy.

      Kinetic energy = (1/2) (mass) (speed)²

   = (1/2) (20kg) (5 m/sec)²

   = 50 joules .

If the ball is not carrying solar panels or a jet engine, then it can
never have any MORE energy than that. And if, as we always
assume, it doesn't suffer any air resistance, then it won't LOSE
any energy. It has 50 joules of energy, permanently or until it hits
something.

If you toss it straight up, then it keeps climbing until it runs out of
kinetic energy. That is, its kinetic energy gets converted to potential
energy as it goes higher and slower. At the top, where it stops
momentarily and has no kinetic energy at all, all the energy it has
is potential energy ... the entire 50 joules you gave it with your toss.

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When two point charges are a distance d apart, the magnitude of the electrostatic force between them is F. If the distance betwe

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Answer:

The magnitude of the electrostatic force decreases by a factor 9

Explanation:

The electrostatic force between two charges is given by:

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where

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d is the distance between the two charges

We see that the magnitude of the force is inversely proportional to the square of the distance. If the distance is increased to 3d: d' = 3d, the new electrostatic force would be:

$F'=k\frac{q_1 q_2}{(d')^2}=k\frac{q_1 q_2}{(3d)^2}=\frac{1}{9} k\frac{q_1 q_2}{d^2}=\frac{F}{9}$