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Anastasy [175]
3 years ago
14

With no friction, you can use the relationship between potential and kinetic energy to predict the speed of the car at the botto

m of this hill from its starting height. To do this start by setting the kinetic and potential energy equations equal to one another
Physics
1 answer:
Dvinal [7]3 years ago
3 0

The speed of the car at the bottom of the hill is obtained as, v = \sqrt{2gh}

According the principle of conservation of energy, the total potential energy of the car will be converted to maximum kinetic energy when the car is at the bottom of the hill.

K.E = P.E\\\\\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v= \sqrt{2gh}

where;

  • <em>v </em><em>is the speed of the car at the bottom of the hill</em>
  • <em>h </em><em>is the height of the hill</em>
  • <em>g </em><em>is acceleration due to gravity</em>

Thus, the speed of the car at the bottom of the hill is obtained as, v = \sqrt{2gh}

Learn more about conservation mechanical energy here: brainly.com/question/332163

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Answer:

Incomplete question

Check attachment for the given diagram

Explanation:

Given that,

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Distance travelled before coming to rest is 6m

Since it comes to rest, then, the final velocity is 0m/s

v=3m/s

Using equation of motion to calculate the linear acceleration or tangential acceleration

v²=u²+2as

0²=3²+2×a×6

0=9+12a

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Then, a=-9/12

a=-0.75m/s²

The negative sign shows that the cylinder is decelerating.

Then, a=0.75m/s²

So, using the relationship between linear acceleration and angular acceleration.

a=αr

Where

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And r is radius

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From the diagram r=250mm=0.25m

Then,

α=0.75/0.25

α =3rad/sec²

The angular acceleration is =3rad/s²

b. Time take to come to rest

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What is the momentum of a 35–kilogram cart moving at a speed of 1.2 meters/second?
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Given:-

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To Find: Momentum of the cart.

We know,

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