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2 years ago

What happens to heat energy after an object is cooled down lolololol asking for my bff

Physics

1 answer:

alisha [4.7K]2 years ago

7 0

Answer:

Heat energy does not cool. Objects cool. Energy is not an object. Now, what happens when an object cools is that it gives off some of its thermal energy by one or more mechanisms: radiation, conduction, or convection. In radiation the energy escaped the object as electromagnetic waves - you see the red hot poker slowly dim as it cools, in conduction the energy is transferred Mechanically by the atoms of the hot object being in physical contact with those of the cool object and in convection the hot material moves to a cooler region where it gives up its heat energy by either radiation or conduction.

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A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the

Mashutka [201]

Answer:

4v/3

Explanation:

Assume elastic collision by the law of momentum conservation:

$m_1v = m_1v_1 + m_2v_2$

where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively

Substitute $m_2 = m_1/2 \& v_1 = v/3$

$m_1v = \frac{m_1v}{3} + \frac{m_1v_2}{2}$

Divide both side by $m_1$ , then multiply by 6 we have

$6v = 2v + 3v_2$

$3v_2 = 4v$

$v_2 = \frac{4v}{3}$

So the final speed of the second car is 4/3 of the first car original speed

5 0

2 years ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

3 0

3 years ago

{WILL GIVE (Brain lies t) TO FULL ANSWERS}

kap26 [50]

Answer:

1: surface temperature

2: red giant

3: The brightest stars are called supergiants. Star clusters are rich in stars just off the main sequence called red giants. Main sequence stars are called dwarfs.

4: A white dwarf is very dense

5: red giant

plz mark me as brainliest :)

4 0

2 years ago

Read 2 more answers

State two applications of electrostatic

kaheart [24]

Atmospheric electricity and storms, electrostatic control filters, and industrial electrostatic seperation as well as spark discharge. these are just a few. hope it helps.

6 0

2 years ago

What is the acceleration of a Porsche that can go from 15 mi/hr to 75 mi/hr in 4 seconds?

Elodia [21]

Hi there!

Acceleration = change in velocity / change in time = Δv/Δt

Thus:

a = (75 - 15)/4 = 60/4 = 15 mi/hr²

8 0

2 years ago

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