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3 years ago

10

Which of the following graphs best represents the relationship between the gravitational potential energy of a freely falling ob

ject and the object's height above the ground? Group of answer choices

1 answer:

VLD [36.1K]3 years ago

3 0

Answer:

The last graph.

Explanation:

Gravitational potential energy is the energy possessed by a body at a given height from the Earth's surface.

The formula to find the gravitational potential energy is given as:

$U=mgh$

Where, 'U' is the gravitational potential energy.

'm' is the mass of the body.

'g' is the acceleration of the body due to gravity.

'h' is the height of the body above the Earth's surface.

So, from the above equation, it is clear that, gravitational potential energy is directly proportional to the height. So, as height increases, the gravitational potential energy increases. At the surface of Earth, where, height is 0, the gravitational potential energy is also zero.

Therefore, the correct graph is a straight line with positive slope and passing through the origin. So, the last option is the correct one.

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Answer:

i.

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As we know

$F_{f}=u*F_{N}=u*m*g$

Also the center ward direction forces

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$a_{c}=\frac{v_{t}^2}{r}$

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But now vt relation with the tangential acceleration

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replacing

$F_{fc}=m*a_{t}*\frac{2\pi*r}{2r}$

$F_{fc}=m*a_{t}*\pi$

So magnitude of the force can get by

$F_{f}=\sqrt{(m*a_{t}*\pi)^{2}+(m*a_{t})^{2}}$

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3 years ago

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Serggg [28]

Answer:

(a) α = 35.20 rad/s^2

(b) θ = 802°

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(d) a = 156.64 cm/s^2

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You replace the values of the parameters in the equation (1):

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The angular acceleration of the disc, for the given time, is 35.20rad/s^2

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In degrees you have:

$\theta=14.00rad*\frac{180\°}{\pi \ rad}=802\°$

The angle described by the disc is 802°

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$v=\omega r$ (3)

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The tangential speed is 139.73 cm/s

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$a=\alpha r$

α: angular acceleration for t=0.892s

$a=(35.20rad/s^2)(4.45cm)=156.64\frac{cm}{s^2}$

The tangential acceleration is 156.64cm/s^2

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