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Illusion [34]
3 years ago
9

Which objects have both potential and kinetic energy? Check all that apply.

Physics
2 answers:
wariber [46]3 years ago
3 0
A falling raindrop
Kinetic energy and potential energy are both applied when a body or object is falling.
denis23 [38]3 years ago
3 0

Answer:

a sky diver who is midway through his descent to the ground

a roller coaster car halfway down the second hill

a falling raindrop

Explanation:

From the law of conservation of energy, it can neither be created nor destroyed but can be converted from one form to another.

The sum of kinetic energy and potential energy is mechanical energy. A body has potential energy due to its position or configuration and kinetic energy due to its speed.

A body can have both potential energy and kinetic energy.

The correct options are:

  • a sky diver who is midway through his descent to the ground
  • a roller coaster car halfway down the second hill
  • a falling raindrop

A boulder resting on top of a mountain  has just potential energy.

A parked car  also has just potential energy.

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What is the acceleration if we speed up from 10m/s to 30 m/s in 10 seconds
lina2011 [118]
Acceleration= change in speed/ change in time

30-10= 20 (change in speed)
Time= 10 seconds

20/10= 2
Acceleration= 2 m/s^2

Hope this helps! :)
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3 years ago
How does the ratio of tin to copper affect the properties of the alloy bronze?
Pani-rosa [81]

Answer:

Stronger and harder than either of the pure metals

Explanation:

7 0
2 years ago
Convert the following measurements as indicated, show work. write the answer in scientific notation.
mojhsa [17]

Answer:

A. 0.95 m

B. 1,100 ml

C. 17 km

D. 500,000 g

Explanation:

A. 95/100

B. 1.1 x 1,000

C. 17,000/1,000

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4 0
3 years ago
Read 2 more answers
A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius
chubhunter [2.5K]

Answer:

a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2

b) See the picture

Explanation:

We can use Gauss's law to find the electric field in all the regions:

EA = qen/e0 where qen is the enclosed charge

Remember that the electric field everywhere outside a sphere is:

E(r) = q/(4*pi*eo*r^2) = Kq/r^2

a)

  1. For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0                       EA = 0/e0 = 0                                                                                                    E = 0
  2. For R < r < 2R: Here the enclosed charge is equal Q                                      E =  Q/(4*pi*eo*r^2) = KQ/r^2      
  3. For r > 2R: Here the enclosed charge is equal 2Q                                              E =  Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2

b)  At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance

7 0
3 years ago
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