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adelina 88 [10]
3 years ago
12

A 500g object falls off a cliff and losers 100 J from its gravitational potential energy store. if the gravitational field stren

gth g=9.8N/kg, how high is the cliff?
Physics
1 answer:
ludmilkaskok [199]3 years ago
7 0

Answer : Height, h = 20.4 m

Explanation :

It is given that,

Mass of an object, m = 500 g = 0.5 kg

Gravitational potential energy, PE = 100 J

The Gravitational potential energy is the energy which is possessed due to the height and gravity of an object. It is given as :

PE = m g h

where,

h is the height of the cliff.

100\ J=0.5\ kg\times 9.8\ m/s^2\times h

h = 20.40 m

So, the height of the cliff is 20.4 m.

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what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates wa
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Answer:

Final energy = Uf = initial energy × d₂/d₁

Explanation:

Energy is the ability to do work.

capacitor is an electronic device that store charges

where

V is the potential difference

d is the distance of seperation between the two plates

ε₀ is the dielectric constant of the material used in seperating the two plates, e.g., paper, mica, glass etc.

A = cross sectional area

U =¹/₂CV²

C =ε₀A/d

C × d=ε₀A=constant

C₂d₂=C₁d₁

C₂=C₁d₁/d₂

charge will  'q' remains same in the capacitor, if the capacitor was disconnected from the electric potential source (v) before the separation of the plates was replaced

Energy=U =(1/2)q²/C

U₂C₂ = U₁C₁

U₂ =U₁C₁ /C₂

U₂ =U₁d₂/d₁

Final energy = Uf = initial energy × d₂/d₁

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What is the magnitude of the gravitational force exerted by earth on a 9.0-kg brick when the brick is in free fall?
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The magnitude is 9.000kg
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Why is a pencil bent when bent in water
wel

Answer:

Refraction

Explanation:

A pencil bends when it enters the water media due to the phenomenon of refraction.

  • Refraction is one of the properties of waves.
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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
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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

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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.

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