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Zarrin [17]
2 years ago
8

In the opposite figure, the work done by the force F to push the box upwards from the ground to the top of the inclined plane is

.......... a) 300 J c) 600 J (b) 450 J d) 750 J F=100 N CamScanner Wa 3m In the opposite figure , the work done by the force F to push the box upwards from the ground to the top of the inclined plane is .......... a ) 300 J c ) 600 J ( b ) 450 J d ) 750 J F = 100 N CamScanner Wa 3m​

Physics
1 answer:
goldfiish [28.3K]2 years ago
4 0

In the opposite figure, the work done by the force F to push the box upwards from the ground to the top of the inclined plane is 600 J.Option c is correct.

<h3>What is work done?</h3>

Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

The displacement value from the triangle is;

\rm sin 30^0 = \frac{P}{H} \\\\ H=S =\frac{3 \ m}{sin 30^0} \\\\ S = 6\ m

The work done is found as ;

\rm W= Fd \\\\ W = 100  \ N \times 6 \\\\ W= 600 \ J

The force F in the opposing figure does 600 J of effort to lift the box from the ground to the top of the inclined plane.

Hence option c is correct.

To learn more about the work done refer to the link;

brainly.com/question/3902440

#SPJ1

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Answer:

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Explanation:

The equations of the position and velocity of the antelope is given by the following expressions:

x = x0 + v0 · t + 1/2 · a ·t²

v = v0 + a · t

Where:

x = position of the antelope at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

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x = x0 + v0 · t + 1/2 · a ·t²

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

We also know that at the second point the velocity is 15.7 m/s. Then at t = 7.50 the velocity will be 15.7 m/s.

v = v0 + a · t

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a = (15.7 m/s - v0) / 7.50 s

Replacing it in the equation for position:

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

68.0 m = v0 · 7.50 s + 1/2 · (15.7 m/s - v0) / 7.50 s · (7.50 s)²

68.0 m = v0 · 7.50 s + 7.85 m/s · 7.50 s - 3.75 s · v0

68.0 m - 7.85 m/s · 7.50 s = 3.75 s · v0

(68.0 m - 7.85 m/s · 7.50 s) / 3.75 s = v0

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The speed of the antelope at the first point is 2.43 m/s.

B) The acceleration of the antelope will be:

a = (15.7 m/s - v0) / 7.50 s

a = (15.7 m/s - 2.43 m/s) / 7.50 s

a = 1.77 m/s²

The acceleration of the antelope is 1.77 m/s²

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