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2 years ago

A small car is pushing a large truck. They are speeding up.

Physics

1 answer:

Gnom [1K]2 years ago

3 0

The force of the truck on the car is equal to the force of the car on the truck.

<h3>What is Newton's third law of motion?</h3>

Newton's third law of motion states that action and reaction are equal and opposite.

Let the mass of the small car = m

Let the mass of the truck = M

The force exerted by the small car is calculated as;

F₁ = ma

The force exerted by the large truck is calculated as;

F₂ = Ma

According to Newton's third law, the magnitude of the two forces are equal but opposite in direction.

|F₁| = |F₂|

Thus, we can conclude that, the force of the truck on the car is equal to the force of the car on the truck.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

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Answer:

Explanation:

From the work-energy theorem, the workdone by a force on a body causes a change in kinetic energy of the body.

But, remember that the work done (W) by a force (F) on a body is the product of the force and the distance d, moved by the body caused by the force. i.e

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3 years ago

Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass g

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Answer:

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Explanation:

Given:

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3 years ago

You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,

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Answer:

The value is $T_c = 12 .1 ^oC$

Explanation:

From the question we are told that

The mass of the ice cube is $m_i = 7.50 *10^{-3} \ kg$

The temperature of the ice cube is $T_i = 0^o C$

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The final temperature of both substance is $T_f = 0^oC$

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

$Q = m_c * c_c * [T_c - T_f ]$

The specific heat of copper is $c_c = 385J/kg \cdot ^oC$

Generally the heat gained by the ice cube is mathematically represented as

$Q_1 = m_i * L$

Here L is the latent heat of fusion of the ice with value $L = 3.34 * 10^{5} J/kg$

$Q_1 = 7.50 *10^{-3} * 3.34 * 10^{5}$

=> $Q_1 = 2505 \ J$

$2505 = 0.540 * 385 * [T_c - 0 ]$

=> $T_c = 12 .1 ^oC$

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Answer:

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