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damaskus [11]
2 years ago
8

an object has a mass of 50kg, a final height of 20m and an initial height of 8m. what is the amount of work done

Physics
1 answer:
Andrei [34K]2 years ago
7 0

amount of work done is 5880 J

Given:

mass of object = 50kg

Final height = 20m

initial height = 8m

To Find:

amount of work done

Solution:

work is done when a force acts upon an object to cause a displacement. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

The work done by gravity is given by the formula,

W = mgh

W = 50 x 9.8 x ( 20-8)

= 5880 J

So the work done is 5880 J

Learn more about Work done here:

brainly.com/question/25239010

#SPJ4

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4 years ago
compare 51.5 hectograms to 51500 decigrams. How do each of these values compare to a gram, and which represents a larger mass?
Olin [163]
These values when compared seem to be the same. They are equal. If we convert them to the same units, they results to the value which is:
 
51.5 Hectograms = 5100 Grams

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Hope this answers the question. Have  a nice day.

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3 years ago
An ipod uses a standard 1.7 volt battery. How much resistance is in the circuit if it uses a current of 0.03 amps​
natali 33 [55]

Answer:

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Explanation:

150 is the answer:)

7 0
3 years ago
You are conducting an experiment inside an elevator that can move in a vertical shaft. A load is hung vertically from the ceilin
Ivan

Answer: The elevator must be accelerating.

Explanation:

As the tension force is opposing to the the force of gravity on the load which is hung vertically, and the tension force can adopt any value in order to comply with Newton's 2nd law, if the tension force is less than the force due to gravity, this means that all system is not in equilibrium, so it must be accelerating.

If we assume that the downward is the positive direction, we can write:

mg - T = ma

If T = 0.9 mg, ⇒ mg (1-0.9) =0.1 mg = m a ⇒a = 0.1 g , in downward direction.

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3 years ago
A friend of yours is loudly singing a single note at 412 Hz while racing toward you at 23.0 m/s on a day when the speed of sound
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Doppler Effect is required to solve the problem. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. The general formula of observed frequency, due to Doppler's effect is as follows,

f_0 = (\frac{v+v_0}{v-v_s})f_s

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v = Velocity of Sound

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PART A) Replacing our values we have that,

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4 0
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