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Rasek [7]
3 years ago
14

Explain the law of conservation of energy. Give a specific example using kinetic and potential energy that shows how energy is c

onserved.
I already know what the law of conservation of energy is, but I need some help coming up with an example. Any ideas or help would be greatly appreciated.
Physics
1 answer:
Ede4ka [16]3 years ago
7 0

Answer:

As you said you already know, energy cannot be created or destroyed.

Explanation:

You cannot gain energy or lose energy, it can only be converted. So if you start on a 3m high hill and go down it, your potential energy is equal to mgh, and if you get to the bottom of the hill, your KE would be equal to your PE at the top, and when you start going up another hill again, the maximum height you can reach is 3m, because energy cannot be created or destroyed, and your mass and gravitational acceleration are the same, so therefore you can only reach the same height you started from due to the conservation of energy.

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a 10kg model rocket is fired at a 60 degree angle from horizontal from the football field with a thrust of 200N. what is the acc
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Answer:

the acceleration of the rocket is: a=vemΔmΔt−g a = v e m Δ m Δ t − g .

Explanation:

I answered this before.

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5 0
2 years ago
A cheerleader lifts his 79.4 kg partner straight up off the ground a distance of 0.945 m before releasing her. the acceleration
Oksi-84 [34.3K]
To find out how much work he has done, we must first calculate force using the force formula (F= Mass*Acceleration). In this case, mass is 79.4 and acceleration is the gravitational constant of 9.8m/s, plugging this into the formula we find that force is 778.12Newtons. Next, we need to multiply force by the distance to get the amount of energy used to lift his partner once. Which is 778.12 * .945 = 735.32. Finally, we need to multiply 735.32 by the number of times he lifts his partner, 33, to get 735.32 * 33 to find that the energy he has expended 24,265.56 Joules of energy.
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2 years ago
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marta [7]
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7 0
3 years ago
Read 2 more answers
A projectile of mass 1.800 kg approaches a stationary target body at 4.800 m/s. The projectile is deflected through an angle of
Artyom0805 [142]

Answer:

P = 5.22 Kg.m/s

Explanation:

given,

mass of the projectile = 1.8 Kg

speed of the target = 4.8 m/s

angle of deflection = 60°

Speed after collision = 2.9 m/s

magnitude of momentum after collision = ?

initial momentum of the body = m x v

                                                  = 1.8 x 4.8 = 8.64 kg.m/s

final momentum after collision

momentum along x-direction

P_x = m v cos θ

P_x = 1.8 x 2.9 x  cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x  sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

P = \sqrt{P_x^2 + P_y^2}

P = \sqrt{4.52^2 + 2.61^2}

 P = 5.22 Kg.m/s

momentum magnitude after collision is equal to P = 5.22 Kg.m/s

5 0
3 years ago
Enter an expression in the box to Write the equation of the line perpendicular to y=-3x-1 that passes through the point (3,4).
Dafna11 [192]

Answer: -3x+13

Explanation:

4 0
2 years ago
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