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Mazyrski [523]
3 years ago
7

A roller coaster car has 600,00 J of Kinetic energy as it approaches the station to stop. The roller coaster comes to a complete

stop. How much work did the brakes do in stopping the car?
Physics
1 answer:
guajiro [1.7K]3 years ago
7 0

Answer:

Work Done by the Brakes = - 60000 J

Negative sign indicates the opposite direction of force and displacement.

Explanation:

Assuming that frictional effects are negligible, we can say that all the work done by the brakes is used to stop the car. We can apply law of conservation of energy to this situation as follows:

Work Done by the Brakes = Change in Kinetic Energy

Work Done by the Brakes = Final Kinetic Energy - Initial Kinetic Energy

Final Kinetic Energy of the roller coaster will be 0. Because, the roller coaster will stop finally and its velocity will become zero.

Work Done by the Brakes = 0 J - 60000 J

<u>Work Done by the Brakes = - 60000 J</u>

<u>Negative sign indicates the opposite direction of force and displacement.</u>

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The benefits of jumping rope include: (which of these things)
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Answer:

E.  All of the Above

Explanation:

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3 0
2 years ago
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mart [117]
A is the correct answer
5 0
3 years ago
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Which factor indicates the amount of charge on the source charge?
Westkost [7]

Answer:

B. the number of field lines on the source charge

Explanation:

As we know that electric flux is defined as the number of electric field lines passing through a given area.

So here  electric flux due to a point charge "q" is given by

so here we know that flux depends on the magnitude of charge and hence we can say that number of filed lines originating from a point charge will depends on the magnitude of the charge.

3 0
3 years ago
The Millersburg Ferry (m = 13000.0 kg loaded) is travelling at 11 m/s when the engines are put in reverse. The engineproduces a
Pie

Explanation:

It is given that,

Mass of Millersburg Ferry, m = 13000 kg

Velocity, v = 11 m/s

Applied force, F = 10⁶ N

Time period, t = 20 seconds

(a) Impulse is given by the product of force and time taken i.e.

J=F.\Delta t

J=10^6\ N\times 20\ s

J=2\times 10^7\ N-s

(b) Impulse is also given by the change in momentum i.e.

J=\Delta p=p_f-p_i

J=p_f-p_i

p_f=J+p_i

p_f=2\times 10^7\ N-s+13000\ kg\times 11\ m/s

p_f=20143000\ kg-m/s

(c) For new velocity,

v_f=\dfrac{p_f}{m}

v_f=\dfrac{20143000\ kg-m/s}{13000\ kg}

v_f=1549.46\ m/s

Hence, this is the required solution.

3 0
3 years ago
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Please help me with this question, especially the angle part. I have solved the initial velocity already. I don't know what to d
Helga [31]
I think you almost got it.

At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.

With the value of the initial speed (28 m/s, which is the total speed), you can set

v_x = v * cos( theta ) ---> 12 = 28*cos(theta) --> cos(theta)=12/28=3/7

or theta = 64.62 deg, it is D. Think about it. I hope you see it.
5 0
2 years ago
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