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Ede4ka [16]
3 years ago
15

PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton b

ox 0.8 meters. 2. A boy lifts a 5-newton box 0.8 meters. 3.A boy lifts a 8-newton box 0.2 meters. 4.A boy lifts a 10-newton box 0.2 meters.
Physics
1 answer:
GuDViN [60]3 years ago
3 0

Explanation:

Work done is given by the product of force and displacement.

Case 1,

1. A boy lifts a 2-newton box 0.8 meters.

W = 2 N × 0.8 m = 1.6 J

2. A boy lifts a 5-newton box 0.8 meters.

W = 5 N × 0.8 m = 4 J

3. A boy lifts a 8-newton box 0.2 meters.

W = 8 N × 0.2 m = 1.6 J

4. A boy lifts a 10-newton box 0.2 meters.

W = 10 N × 0.2 m = 2 J

Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.

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A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin
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Answer:

a) \omega \approx 219.911\,\frac{rad}{s}, b) \alpha = 16.916\,\frac{rad}{s^{2}}, c) a_{t} = 1.776\,\frac{m}{s^{2}}, d) a_{n} = 5077.889\,\frac{m}{s^{2}}, e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation, f) Zero, g) v = 23.091\,\frac{m}{s}.

Explanation:

a) The maximum angular velocity of the fan is:

\omega = (35\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )

\omega \approx 219.911\,\frac{rad}{s}

b) The angular acceleration of the fan is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{219.911\,\frac{rad}{s}-0\,\frac{rad}{s}}{13\,s}

\alpha = 16.916\,\frac{rad}{s^{2}}

c) The magnitude of the tangential aceleration is:

a_{t} = (16.916\,\frac{rad}{s^{2}} )\cdot (0.105\,m)

a_{t} = 1.776\,\frac{m}{s^{2}}

d) The magnitude of the centripetal acceleration is:

a_{n} = (219.911\,\frac{rad}{s} )^{2}\cdot (0.105\,m)

a_{n} = 5077.889\,\frac{m}{s^{2}}

e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation.

f) When fan is at full speed, it rotates at constant rate and, hence, there is no angular acceleration. Besides, the tangential acceleration experimented by the gum is zero.

g) The linear speed of the gum is:

v = (219.911\,\frac{rad}{s} )\cdot (0.105\,m)

v = 23.091\,\frac{m}{s}

5 0
3 years ago
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