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PilotLPTM [1.2K]
3 years ago
5

Which example shows potential energy? A. a skydiver falling B. a car racing C. hitting a nail with a hammer D. a wound up watch

spring
Physics
2 answers:
goldfiish [28.3K]3 years ago
8 0
The answer is D, because all of the others show an object at work, when an object that has potential energy has to be at rest, like when you are at the top of a hill, or the highest point a rocket can go before it comes back done to the ground.
tankabanditka [31]3 years ago
6 0
B.a car racing down a hill
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A traditional unit of length in Japan is the ken (1 ken 1.97 m). What are the ratios of (a) square kens to square meters and (b)
Roman55 [17]
If 1 ken is 1.97 meter, then 1 square ken is 3.8809 square meters, and one cubic ken is 7.645373. As for the cylindrical tank, the volume of it would be 10.835 times the radius of the cylinder time 1.97^2 times pi. As you didn't specify the radius, I can't give the exact answer but that would be how to get it.
7 0
3 years ago
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A container is filled with water to a depth of 26.2 cm. On top of the water floats a 16.1 cm thick layer of oil with a density of
Solnce55 [7]

Answer:

1.34 x 10^3 Pa

Explanation:

density of oil = 0.85 x 10^3 kg/m^3

g = 9.81 m/s^2

height of oil column = 16.1 cm = 0.161 m

Pressure on the surface of water = height of oil column x density of oil x g

                                                      = 0.161 x 0.85 x 10^3 x 9.81 = 1.34 x 10^3 Pa

Thus, the pressure on the surface of water is 1.34 x 10^3 Pa.

4 0
3 years ago
A balloon is expanded to the same volume as that of a human head. Do an order-of-magnitude estimate of the volume of this balloo
cestrela7 [59]

Answer:

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Explanation:

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 Volume of sphere =\frac{4}{3} \pi r^3, where r is the radius.

 We have approximate radius = 10 cm.

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 In the given options the closest value to the approximate volume is 1000 cm^3.

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4 0
3 years ago
HELLO , PLZ HELP .
sergey [27]

Answer:

the pressure due to the water on the diver is 200,000 pascal

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p=200,000 pascal

6 0
3 years ago
A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (
natita [175]

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

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b )

When power is switched off , it will decelerate because of frictional torque .

5 0
3 years ago
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