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Anastaziya [24]
3 years ago
9

What happens when a roller coaster car moves down from the top of a hill?

Physics
2 answers:
Andrew [12]3 years ago
5 0

Answer:

Gravitational potential energy is converted into kinetic energy

Explanation:

At the top of the hill, the roller coaster car has gravitational potential energy, which is given by:

U=mgh

where m is the mass of the car, g is the gravitational acceleration, h is the height of the hill.

As the car descends the hill, it gains speed (v), so it also gains kinetic energy, given by:

K=\frac{1}{2}mv^2

where v is the speed of the car.

The mechanical energy of the car is the sum of its potential energy and kinetic energy:

E=U+K

In absence of friction, the law of conservation of energy states that the mechanical energy of the car is constant. Therefore, since as the car moves down the hill the potential energy decreases (because the height, h, decreases), it means that the kinetic energy must increase (and this is wy the speed increases). So we can say that gravitational potential energy is converted into kinetic energy.

VLD [36.1K]3 years ago
3 0
When the roller coaster moves down from the top of the hill, all of its stored potential energy is converted into kinetic energy to move it and when it goes back up the hill it turns kinetic into potential.

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natita [175]

Answer:

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Explanation:

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Explain how fluids exert pressure
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4 0
3 years ago
The peak value of an alternating current in a 1500-W device is 6.4 A. What is the rms voltage across it
horrorfan [7]

Answer:

6787.5 V

Explanation:

From the question,

P = IV..................... Equation 1

Where P = Power, I = rms current, V = rms voltage.

make V the subject of the equation

V = P/I................. Equation 2

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Substitute these values into equation 2

V = 1500(4.525)

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Which process has led to the production of most modern domesticated crops and livestock
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6 0
3 years ago
Read 2 more answers
A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees.Find her final speed at th
Shalnov [3]
Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.= 
<span>Wt. of skier. </span>

<span>Fp=588*sin35 = 337 N.=Force parallel to </span>
<span>incline. </span>
<span>Fv = 588*cos35 = 482 N. = Force perpendicular to incline. </span>

<span>Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force </span>
<span>of kinetic friction. </span>
<span>d =h/sinA = 2.5/sin35 = 4.36 m. </span>

<span>Ek + Ep = Ekmax - Fk*d </span>
<span>Ek = Ekmax-Ep-Fk*d </span>
<span>Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J. </span>
<span>Ek = 0.5m*V^2 = 2682 J. </span>
<span>30*V^2 = 2682 </span>
<span>V^2 = 89.4 </span>
<span>V = 9.5 m/s = Final velocity.</span>
4 0
4 years ago
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