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LekaFEV [45]
3 years ago
9

The transfer of heat between objects that are touching is called what?

Physics
2 answers:
Ivenika [448]3 years ago
7 0

Answer:

conduction if i hold a hot cup the cup is giving heat which comes in contact with my hands warming my hands

Savatey [412]3 years ago
6 0
Conduction is heat transfer between two objects that are touching each other. When two objects touch and one has a higher temperature than the other; heat is transferred to the object with the lower temperature. When you touch something hot it feels hot because heat is being transferred from the object to your hand.
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The strong nuclear force holds the nucleus of an atom together.  

Somehow, it overcomes the electrical force of repulsion between protons in the nucleus, which all have the same charge but still stay close together somehow. (b)

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A hoop and a solid disc are relased from rest
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Answer:

1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

<u>For the hoop:</u>

K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

I_h = m_hR^2

Let's continue with the energy equations:

m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}

Similarly <u>for the solid disk</u> with a moment of inertia of (1/2)mR^2:

K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2

where a is the linear acceleration.

The relation with the angular and linear acceleration is

a = \alpha R

where R is the radius of the tire. Since it is not given in the question, we will leave it as R.

The angular acceleration of the small pebble is

\alpha = 2.235/R ~m/s^2

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GEL IT IS COMBINATION OF SOLID AND LIQUID
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