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LekaFEV [45]
3 years ago
14

When a rock is held above the ground, we say it has some potential energy. When we let it go, it falls and we say the potential

energy is converted to kinetic energy. Finally, the rock hits the ground (and stays there). What has happened to the energy?
Physics
1 answer:
snow_lady [41]3 years ago
8 0

Answer:

Energy converted to heat and sound. Deformation of soil or ground may also take up some energy.

Explanation:

When the rock hits the ground, a part of the energy is converted to heat (thermal energy) and sound. If the ground is soft, it will be compressed by the rock when it lands. This deformation of the ground will take up some of the energy.

We can think of kinetic energy being transferred to the ground in the following way:

1. Ball hits the ground, and transfers kinetic energy to lots of soil particles.

2. Soil particles move a bit and transfer the energy to soil particles ahead of them.

3. Kinetic energy is transferred to a very large number of soil particles, and they move so little that we consider this vibration motion.

4. This small vibration motion is now considered as internal energy or 'heat'.

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Practice questions, will mark brainliest!
andrew-mc [135]

Answer:

266.67Watts

Explanation:

Time = 2.5hr to seconds

3600s = 1hr

2.5hrs = 3600×2.5= 9000s

Force = 32N

Distance = 75km  to m

1000m = 1km

75km = 1000×75 = 75000m

Power = workdone / time

Work = force × distance

Therefore work = 32N × 75000m

Work = 2400000Nm

Power = work ➗ time

Power = 2400000Nm ➗ 9000s

Power = 266.67Watts

Watts is the S. i unit of power

I hope this was helpful, please mark as brainliest

4 0
3 years ago
What magnification would be obtained if an eyepiece with a focal length of 0.38 m was placed on telescope?
weqwewe [10]

Answer:

This question is incomplete

Explanation:

This question is incomplete because the telescope's focal length was not provided. The formula to be used here is

Magnification = telescope's focal length/eyepiece's focal length

The eyepiece's focal length was provided in the question as 0.38 m.

NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).

7 0
3 years ago
Why the specific heat capacity of the sun remain constant<br><br>​
gayaneshka [121]
The heat remains constant because there’s nothing to cool it down
7 0
3 years ago
A satellite’s velocity is 30000m/s. After 60 secs, it’s velocity shows to 15000m/a. What is the satellite’s acceleration?
Orlov [11]

Answer:

Acceleration = 9 × 10^5 m/s^2 ( deceleration )

Explanation:

From the first equation of motion:

V = u + at

15000 = 30000 + 60a

a = ( 15000-30000)/60

a = 9 × 10^5 m/s^2

4 0
3 years ago
at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g
Readme [11.4K]

The mass of water that must be raised is 5.25\cdot 10^7 kg

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

P_{out} = 0.70 P_{in}

where P_{in} is the power in input.

The power in input can be written as

P_{in} = \frac{W}{t}

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

W=mgh

where

m is the mass of water

g=9.8 m/s^2 is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

P_{out} = 0.70 \frac{mgh}{t}

Where

P_{out} = 150 MW = 150\cdot 10^6 W

And solving for m, we find:

m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

3 0
3 years ago
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