All categories

3 years ago

Which heat transfer process is important in the transfer of energy from the Sun to Earth?

Physics

1 answer:

kondor19780726 [428]3 years ago

4 0

Answer: Radiation

Explanation:

The heat transfer process that is important in the transfer of energy from the Sun to Earth is Radiation.

Radiation is one of the mode of heat transfer which transfer energy/heat from one medium to another without any intervening medium. For example, heat from the sun reaches the earth through radiation. The sun hits the earth directly without any medium between the sun itself and the earth.

You might be interested in

The kinetic energy <br> as the roller coaster goes downhil

Step2247 [10]

Answer:

Increases

Explanation:

The kinetic energy _____ (Increses)

as the roller coaster goes downhill

4 0

2 years ago

Read 2 more answers

What is the motion of the particles in this kind of wave?

Musya8 [376]

The answer to this question is B I think

7 0

3 years ago

Read 2 more answers

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

3 years ago

Select the correct answer.

kipiarov [429]

Answer:

Explanation:

If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.

5 0

3 years ago

How did Bourne believe that an object could be made to rise or sink at will?

Angelina_Jolie [31]

Bourne believed that an object would float or sink at will as long as he could <span>manipulate the effect's of buoyancy which control and object to sink or float. Hope this helps!

</span>

4 0

4 years ago

Read 2 more answers