All categories

2 years ago

What type of heat does not require matter?

Physics

2 answers:

koban [17]2 years ago

8 0

Radiation is the only method of heat transfer that does not require matter.<u> </u>

<u>Facts</u><u> </u><u>about</u><u> </u><u>Radiation</u> :

The energy from the sun is received in the form of radiation.
It travels in the form of waves.
It can be absorbed by substances in its path.
Heat Radiation travels through air.
Black surfaces are the emitters of heat radiation.
Shiny surfaces are the reflectors of heat radiation.
Heat transfer by Radiation doesn't require any medium.

Lana71 [14]2 years ago

4 0

It would be Thermal Radiation

You might be interested in

Why is gravity an example of a scientific law?

valina [46]

Because a sxientific law is always applies under the same conditions, and implies that there is a causal relationship involving its elements. And so that is why gravity <span>always applies under the same conditions, and implies that there is a causal relationship involving its elements.</span>

7 0

2 years ago

Read 2 more answers

A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

<u>Mathematical expression for the Newton's second law of motion is given as:</u>

$F=\frac{dp}{dt}$ ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

$p=m.v$

Now, equation (1) becomes:

$F=\frac{d(m.v)}{dt}$

<em>∵mass is constant at speeds v << c (speed of light)</em>

$\therefore F=m.\frac{dv}{dt}$

and, $\frac{dv}{dt} =a$

where: a = acceleration

$\Rightarrow F=m.a$

also

$F\propto \frac{1}{dt}$

so, more the time, lesser the force.

<em>& </em><u><em>Impulse:</em></u>

$I=F.dt$

$I=m.a.dt$

$I=m.\frac{dv}{dt}.dt$

$I=m.dv=dp$

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0

3 years ago

Pls help for brainliest:)

Rasek [7]

Answer:

Explanation:

<h3>#CARRY ON LEARNING</h3><h3>#MARK ON LEARNING</h3><h3>#HELPING HAND</h3>

6 0

2 years ago

Use the drop-down menus to complete the statement.

Nikitich [7]

Classics.

Resistance is equal to relation between voltage and current.

$R = \frac{U}{I}$

If we express current:

$I = \frac{U}{R}$

If current is in fact 0 then one of the quantities either voltage or resistance must be equal to zero. Since resistance cannot be equal 0, because that would violate mathematical law that states that division by zero is undefined the only logical conclusion is voltage.

So the answer should be C voltage and B zero.

Hope this helps!

8 0

3 years ago

Read 2 more answers

The <br> of a wave is the density of the medium’s particles at the compression of the wave.

andrew11 [14]

<span>Amplitude is the correct answer. I hope this helps.</span>

6 0

3 years ago

Read 2 more answers