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2 years ago

a cyclist applies a constant forward force of 20 N to maintain a velocity 2.5 m/s. how much power does the cyclist deliver?

Physics

1 answer:

forsale [732]2 years ago

4 0

The power exerted by the cyclist is determined as 50 W.

<h3>Average power exerted by the cyclist</h3>

The power exerted by the cyclist is calculated as follows;

P = FV

where;

F is the applied force
V is velocity

P = 20 x 2.5

P = 50 W

Thus, the power exerted by the cyclist is determined as 50 W.

Learn more about power here: brainly.com/question/25263760

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1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

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Answer with explanation:

We are given that

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Height, $h_1=1.6 m$

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Horizontal velocity, $v_x=2 m/s$

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a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

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We can write it as follows :

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So, the correct option is (a).

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