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3 years ago

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A book rests on a table. The earth pulls down on the book through the weight force. Which force is the reaction force to the wei

ght force?
1. the force of the Earth on the book
2. the force on the table on the book
3. the force of the Earth on the table
4. the force of the book on the Earth
5. the inertia of the book

1 answer:

Veseljchak [2.6K]3 years ago

7 0

Newton's 3rd law: The force that A exerts on B is equal and opposite (in direction) to the force that B exerts on A.

Substitute the words "the earth" for A and "the book" for B.

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A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.58 m. A child appl

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Answer:

The value is $KE = 259.6 \ J$

Explanation:

From the question we are told that

The weight of the horizontal solid disk is $W = 805 \ N$

The radius of the horizontal solid disk is $r = 1.58 \ m$

The force applied by the child is $F = 49.5 \ N$

The time considered is $t = 2.95 \ s$

Generally the mass of the horizontal solid disk is mathematically represented as

$m_h = \frac{W}{ g}$

=> $m_h = \frac{805}{ 9.8 }$

=> $m_h = 82.14 \ N$

Generally the moment of inertia of the horizontal solid disk is mathematically represented as

$I = \frac{1}{2} * m * r^ 2$

=> $I = \frac{1}{2} * 82.14 * 1.58^ 2$

=> $I = 102.5 \ kg \cdot m^2$

Generally the net torque experienced by the horizontal solid disk is mathematically represented as

$T = I * \alpha = F * r$

=> $\alpha = \frac{ F * r }{ I }$

=> $\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }$

=> $\alpha = 0.7628$

Gnerally from kinematic equation we have that

$w = w_o + \alpha t$

Here $w_o$ is the initial angular velocity velocity of the horizontal solid disk which is $w_o = 0\ rad/s$

So

$w = 0 + 0.7628 * 2.95$

=> $w = 2.2503 \ rad/s$

Generally the kinetic energy is mathematically represented as

$KE = \frac{1}{2} * I * w^2$

=> $KE = \frac{1}{2} * 102.53 * 2.2503 ^2$

=> $KE = 259.6 \ J$

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