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liraira [26]
3 years ago
13

Kamala puts on inline skates and stands facing a wall. When she pushes against the wall, she rolls backward. Why does pushing ha

rder make her roll faster?
-The force of friction between the wheels and the ground decreases.

-The force of the wall pushing on her increases.

-The additional force decreases her weight.

-The extra force causes an increase of the normal force from the ground.
Physics
2 answers:
Veseljchak [2.6K]3 years ago
8 0

Because the force of the wall pushing on her increases.

According to Newton's third law of motion, to every action there is an equal and opposite reaction. When the Kamala pushes the wall, the wall also pushes her. Pushing harder against the wall causes the force of the wall pushing her also increase. Thus, making her roll faster.

Svetach [21]3 years ago
7 0

Answer: Second option "The force of the wall pushing on her increases."

Explanation: This is because of Newton's third law: For every action (force) in nature, there is an equal and opposite reaction.

So when Kamala pushes against the wall, the wall is also pushing against Kamala, and this is what impulses her backward. So if she uses more force, then the wall sill apply more force on her, and then she will roll faster.

The right answer is then the second option: "The force of the wall pushing on her increases."

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In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

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We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

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