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jekas [21]
4 years ago
11

Is a light bulb that is on potential or kinetic?

Physics
2 answers:
Likurg_2 [28]4 years ago
8 0

Answer:

pretty sure its kinetic

Explanation:

puteri [66]4 years ago
4 0

Answer:

Kinetic

Explanation:

The stored chemical potential energy of a battery converts to electrical kinetic energy to transport electricity to a light bulb, which radiates thermal kinetic energy.

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When electrons in an atom in an excited state fall to lower energy levels, energy is?
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Radiated away as electromagnetic radiation.
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3 years ago
A 35 kg trunk is pushed 4.8 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic fric
netineya [11]

Answer:

We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

F_x = F\cos(30^\circ)\\F_y = F\sin(30^\circ)

Only the x-component of the applied horizontal force does work on the trunk.

But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

0 = F\cos(30^\circ) - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0.86F = 35(9.8)(0.5) + 35(9.8)(0.86)(0.19)\\F =264.5N

F_x = 264.5\cos(30^\circ) = 227.5N\\F_y = 264.5\sin(30^\circ) = 132.25N

W_{F_x} = F_xd = 227.5\times 4.8 = 1092J

The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

W_g = -mg\sin(30^\circ)d

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

W_g = -823J

The energy dissipated by the frictional force can be found as follows:

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.

In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.

0 = F - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0 = F - 35(9.8)(0.5) - 35(9.8)(0.86)(0.19)\\0 = F - 171.5 - 56\\F = 227.5N

W_F = Fd = 227.5\times 4.8 = 1092J

W_g = -mg\sin(30^\circ)d = -35(9.8)(0.5)(4.8) = -823J

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Explanation:

As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.

8 0
4 years ago
In an inelastic collision involving an isolated system, the final total momentum isA) more than the initial momentum.B) exactly
damaskus [11]

I think that the answer is C

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3 years ago
Do electromagnetic waves travel through a medium
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C) alternating current .
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B)direct current </span>
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Read 2 more answers
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