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grigory
3 months ago
8

A book rests on the shelf of a bookcase. The reaction force to the force of gravity acting on the book is 1. The force of the sh

elf holding the book up. 2. The force exerted by the book on the earth. 3. The weight of the book. 4. The frictional force between book and shelf. 5. None of these.
Physics
2 answers:
kicyunya [3.2K]3 months ago
8 0

Answer:

the force that the book applies to the earth

serg [3.5K]3 months ago
6 0

Answer:

1. The force applied by the shelf supporting the book.

Explanation:

The free body diagram for the book is represented as follows:

1 - The weight of the book acting downward

2 - The normal force exerted by the shelf upward on the book.

As the book remains stationary, these two forces balance each other, and in accordance with Newton's Third Law, the reactive force equivalent to gravity is opposite and equal to the weight of the book. This reaction force prevents the book from falling off the shelf.

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Answer:

  v = 54.2 m/s

Explanation:

We can utilize conservation of energy to solve this issue.

Initial condition Higher

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Final condition. Lower

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        Em₀ = Em_{f}

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Now let's perform the calculation

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3 months ago
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Yuliya22 [3333]

Response:

The acceleration of car 2 is four times that of car 1.

Rationale:

Centripetal acceleration occurs when an object travels in a circular route. It can be expressed as:

a=\dfrac{v^2}{r}

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Thus,

\dfrac{a_1}{a_2}=\dfrac{v_1^2}{v_2^2}

1 and 2 represent the first and second cars, respectively.

v_2=2v_1

Consequently,

\dfrac{a_1}{a_2}=\dfrac{v_1^2}{(2v_1)^2}\\\\\dfrac{a_1}{a_2}=\dfrac{v_1^2}{4v_1^2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{4}\\\\a_2=4\times a_1

Therefore, car 2's acceleration is four times that of car 1.

4 0
2 months ago
What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
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The equation for simple harmonic motion (SHM) is as follows.
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2 months ago
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The work performed to elevate the box equals 515.03 J.
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1 month ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Yuliya22 [3333]

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

It is known that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v are constant on both sides. Therefore we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

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let the final pressure be P and the temperature T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}.............3

divide equation (2) by equation (3)

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, the left side temperature equals 1.48 times the right side temperature

6 0
2 months ago
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