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aliya0001 [1]
2 years ago
10

A wagon of dog treats (combined mass 55 kg) is rolling at 2.1 m/s. A dog with mass 21 kg dives into the wagon, colliding with ju

st enough momentum to make both stop. If the collision between the dog and the wagon lasts 0.1 s, what is the magnitude of the average force that will be exerted on the dog by the collision with the wagon
Physics
1 answer:
Lunna [17]2 years ago
5 0

Answer:

Explanation:

An impulse results in a change of momentum

If the wagon and dog both stop, they must have had equal and opposite momentums

FΔt = mΔv

F = mΔv/Δt = m(v₁ - v₀)/(t₁ - t₀)

v₁ = t₀ = 0

F = m(v₀)/t₁

F = 55(2.1)/0.1 = 1155 N

We could have also figured the dog's initial velocity and used the dog's mass in the equation as well. Result would be identical.

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It takes 50 seconds for Tyler to do 450 Joules of work. How much power did Tyler use?
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Answer:

<h2>The answer is 9 W</h2>

Explanation:

The power used by an object can be found by using the formula

p =  \frac{w}{t}  \\

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We have the final answer as

<h3>9 W</h3>

Hope this helps you

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How do astronomers think the universe was created?
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3 years ago
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Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the
ale4655 [162]

I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) <em>g</em>

• the upward normal force of the surface, magnitude <em>n₁</em>

• kinetic friction, mag. <em>f₁</em> = 0.30<em>n₁</em>, pointing in the negative horizontal direction

• the contact force of the larger block, mag. <em>c₁</em>, also pointing in the negative horizontal direction

• the applied force, mag. <em>F</em>, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) <em>g</em>, pointing downward

• normal force, mag. <em>n₂</em>, pointing upward

• kinetic friction, mag. <em>f₂</em> = 0.30<em>n₂</em>, pointing in the negative horizontal direction

• contact force from the smaller block, mag. <em>c₂</em>, pointing in the <u>positive</u> horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that <em>c₁</em> = <em>c₂</em>, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

<em>n₂</em> - (3.0 kg) <em>g</em> = 0   ==>   <em>n₂</em> = (3.0 kg) <em>g</em>   ==>   <em>f₂</em> = 0.30 (3.0 kg) <em>g</em>

• net horizontal force:

<em>c₂</em> - <em>f₂</em> = 0   ==>   <em>c₂</em> = 0.30 (3.0 kg) <em>g</em> ≈ 8.8 N

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Answer:

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Explanation:

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