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HACTEHA [7]
3 years ago
7

The pulley system below uses a gasoline engine to raise a drill head up through a smooth drill pipe. The engine provides a const

ant force of 900 N. The drill head has a mass of 80 kg. Find the power that is being consumed by the motor (provided by the gasoline) at the instant the drill head has been raised 5 m., starting from rest. The engine has an efficiency of 0.65.
Physics
1 answer:
polet [3.4K]3 years ago
6 0

NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question

Answer:

Power = 2702.56 W

Explanation:

Let the power consumed be P

Energy expended = E = mgh

height, h = 5 m

E = 80 * 9.8 * 5

E = 3920 J

Power = \frac{Energy}{time}

To calculate the time, t

From F = ma

F = 900 N

900 = 80 a

a = 900/80

a = 11.25 m/s²

From the equation of motion, s = ut + 0.5at^{2}

The drill head starts from rest, u = 0 m/s

5 = 0 * t + (0.5*11.25*t^{2} )\\5 = 5.625t^{2}\\t^{2} = 5/5.625\\t^{2} = 0.889\\t = 0.943 s

Power, P = E/t

P = 3920/0.0.943

P = 4157.79 W

But Efficiency, E = 0.65

P = 0.65 * 4157.79

Power = 2702.56 W

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(i) Effect of colour of light: The critical angle for a pair of media is less for the violet light and more for the red light. Thus the critical angle increases with the increase in wavelength of light.

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According to the question,

μ 1​ sinCR​ =1

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initial speed of the stuntman is given as

v = 28 m/s

angle of inclination is given as

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now the components of the velocity is given as

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here it is given that the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave.

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Answer:

Explanation:

Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.

Energy dissipates in 55Ω resistor is given by V²/R

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E=7.09J

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