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emmasim [6.3K]
4 years ago
10

A child lifts a 5.0-newton toy to a height of 0.50 meters. How much work is done on the toy?

Physics
2 answers:
AleksandrR [38]4 years ago
8 0
You can use the formula

Work = force x distance

therefore, 

W= (5.0)(.50)
W = 2.5 joules

hope this helps :)
natali 33 [55]4 years ago
6 0
Work done=mgh
W=force*height
W=5N*0.50mts
W=2.5joule
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Natalija [7]

Answer:

0.775 m

Explanation:

As the car collides with the bumper, all the kinetic energy of the car (K) is converted into elastic potential energy of the bumper (U):

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where we have

k=2\cdot 10^6 N/m is the spring constant of the bumper

x is the maximum compression of the bumper

m=30\cdot 10^4 kg is the mass of the car

v=2.0 m/s is the speed of the car

Solving for x, we find the maximum compression of the spring:

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(30\cdot 10^4 kg)(2.0 m/s)^2}{2\cdot 10^6 N/m}}=0.775 m

8 0
3 years ago
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wlad13 [49]

Answer:

IMA = 2.5 metres

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Explanation:

The AMA of a machine is referred to as the Actual Mechanical Advantage of a machine, calculated as the ratio of the output to the input force.

The Ideal Mechanical Advantage is the ratio of the input distance to the output distance.

From the diagram, the input distance which is also the distance moved by effort  = 5metres

The load distance (output distance) = 2 metres

IMA = INPUT DISTANCE / OUTPUT DISTANCE

IMA = 5metres / 2 metres = 2.5 meters

Efficiency is the ratio of AMA TO IMA

AMA = 2, IMA = 2.5

EFFICIENCY = AMA / IMA

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EFFICIENCY = 80%

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Igoryamba
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3 years ago
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aksik [14]

Answer:

51.85m/s

Explanation:

Given parameters:

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Time taken  = 0.001s

Unknown:

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To solve this problem, we use Newton's second law of motion:

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v is the final velocity

u is the initial velocity

t is the time taken

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