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lesya [120]
3 years ago
7

Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a heigh

t of 45 cm above the bottom. What is its speed at the top of the loop
Physics
1 answer:
Pachacha [2.7K]3 years ago
5 0

Answer:

v = 1.7 m/s

Explanation:

By applying conservation of energy principle in this situation, we know that:

Loss in Potential Energy of Car = Gain in Kinetic Energy of Car

mgΔh = (1/2)mv²

2gΔh = v²

v = √(2gΔh)

where,

v = velocity of car at top of the loop = ?

g = 9.8 m/s²

Δh = change in height = 45 cm - Diameter of Loop

Δh = 45 cm - 30 cm = 15 cm = 0.15 m

Therefore,

v = √(2)(9.8 m/s²)(0.15 m)

<u>v = 1.7 m/s</u>

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A 3250 N car is pushed a distance of 35 m the power was 11375 J, how long did it take?
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Unknown:

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To solve this problem;

 Power is the rate at which work is done

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          11375  = \frac{123200}{t}  

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If a box is pulled with a force of 100 N at an angle of 25
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The X and Y components of the force are 90.63 Newton and 42.26 Newton respectively.

<u>Given the following data:</u>

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To determine the X and Y components of the force:

<h3>The horizontal component (X) of a force:</h3>

Mathematically, the horizontal component of a force is given by this formula:

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Fx = 90.63 Newton.

<h3>The vertical component (Y) of tensional force:</h3>

Mathematically, the vertical component of a force is given by this formula:

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