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SashulF [63]
2 years ago
5

In Figure 10.13, the author illustrates a book being raised at constant speed. Which of the following statements are true? Selec

t all that apply.
There is no change in the kinetic energy of the book.

The potential energy of the book-Earth system decreases.

The potential energy of the book-Earth system increases.

The potential energy of the book increases.

Physics
2 answers:
sergey [27]2 years ago
7 0

The true statement is that ;there is no change in the kinetic energy of the book and the potential energy of the book-Earth system increases.

We have to note that from the law of conservation of mechanical energy, the sum of the kinetic and potential energy of the book at any point remains constant.

If the book is being raised at constant speed, the the true statement is that ;there is no change in the kinetic energy of the book and the potential energy of the book-Earth system increases.

Learn more about kinetic energy: brainly.com/question/15308590

AnnyKZ [126]2 years ago
3 0

Answer:

There is no change in the kinetic energy of the book and the potential energy of the book-Earth system increases.

Explanation:

This is because velocity is constantly increasing. This supports the idea that the kinetic energy of the book doesn't change and that the potential energy of the book-Earth system increases.

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How does energy transfer by waves differ from energy transfer by moving object
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Energy transfer by waves: two primary modes = (electromagnetic waves, compression/transverse waves propagating through a medium)

1) electromagnetic waves:

Using a particle model for the wave (photons for light), energy transfer is similar to that by discrete moving object -- particles carry the energy from one place to another in the absence of a medium.

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3 years ago
Which of the following represents a concave lens?<br> A. -di<br> B. +di<br> C. -f<br> D. +f
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4 0
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Read 2 more answers
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on
leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

q_1=50\ nC at y=6\ m

q_2=-80\ nC at x=-4\ m

q_3=70\ nC at y=-6\ m

Electric Potential is given by

V=\frac{kQ}{r}

Distance of q_1 from x=8\ m

d_1=\sqrt{6^2+8^2}

d_1=\sqrt{36+64}

d_1=10\ m

similarly d_2=8-(-4)

d_2=12\ m

d_3=\sqrt{(-6)^2+8^2}

d_3=\sqrt{36+64}

d_3=10\ m

Potential at x=8\ m is

V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}

V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]

V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}

V_{net}=9\times 5.33

V_{net}=47.97\approx 48\ V

5 0
3 years ago
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