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Finger [1]
4 years ago
14

Which choice correctly describes what happens during heating?

Physics
1 answer:
Masja [62]4 years ago
7 0

Answer:

Option 4

Explanation:

During heating actually heat transfer takes place from a body at higher temperature to a body at lower temperature and the heat transfer takes place until both attain the same temperature  

Therefore heat transfer depends on the temperature of the systems

Now while comparing the thermal energies of the systems, if both the systems have same mass then the system which is at higher temperature has greater thermal energy when compared to the system which is at lower temperature

So in this case assuming that both the systems have same mass then the energy will leave the system with greater thermal energy and go into the system with less thermal energy as the system with greater thermal energy in this case will be at higher temperature and we are considering this assumption because thermal energy not only depends on temperature but also depends on mass of the system

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a ball is dropped from rest at a height of 45.0 m above the ground. ignore the effects of air resistance. What is the speed of c
NemiM [27]

So, the final velocity of the ball when it is 10.0 m above the ground approximately <u>26.2 m/s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses the principle of final velocity in free fall. Free fall occurs only when an object is dropped (without initial velocity), so the falling object is only affected by the presence of gravity. In general, the final velocity in free fall can be expressed by this equation :

\boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}}

With the following condition :

  • v = final velocity (m/s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

<h3>Problem Solving</h3>

We know that :

  • \sf{h_1} = initial height = 45.0 m
  • \sf{h_2} = final height = 10.0 m
  • g = acceleration of the gravity = 9.8 m/s²

Note :

At this point 10 m above the ground, the object can still complete its movement up to exactly 0 m above the ground.

What was asked :

  • v = final velocity = ... m/s

Step by Step

\sf{v = \sqrt{2 \times g \times \Delta h}}

\sf{v = \sqrt{2 \times g \times (h_1 - h_2)}}

\sf{v = \sqrt{2 \times 9.8 \times (45 - 10)}}

\sf{v = \sqrt{19.6 \times 35}}

\sf{v = \sqrt{686}}

\boxed{\sf{v \approx 26.2 \: m/s}}

<h3>Conclusion</h3>

So, the final velocity of the ball when it is 10.0 m above the ground approximately 26.2 m/s.

<h3>See More :</h3>
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