All categories

3 years ago

How is thermal energy transferred during convection?

Physics

2 answers:

Lerok [7]3 years ago

5 0

Conduction and convection it involves particles.

Ne4ueva [31]3 years ago

3 0

Answer: by movement in fluids

You might be interested in

Describe two examples of how humans use electromagnetic waves.

Alex73 [517]

Electromagnetic waves are used in everyday life. You are looking at your computer screen right now. The light that is coming off of the screen is visible light, a form of electromagnetic radiation. Electromagnetic waves are also used to send information. For example, AM or FM radios are radio waves that transfer sound information to your local radio.

5 0

3 years ago

Read 2 more answers

In the morning, if you walk through a grassy lawn or field, quite often your feet will get very wet. What is the liquid generall

Dima020 [189]

The liquid is called condensation and its generally formed when water vaor in the air cools down rapidly and collects around or on an object forming water droplets.

6 0

3 years ago

Which advantage of stored digital data could be a disadvantage in some

Lesechka [4]

Answer:B- All types of information can be pressed on to receivers of the data.

Explanation: hope this helps

4 0

3 years ago

Determine the distance between a newly discovered planet and its single moon if the orbital period of the moon is 1.2 Earth days

Vinil7 [7]

Answer:

The distance is $r = 55430496 \ m$

Explanation:

From the question we are told that

The period of the moon $T = 1.2 days = 1.2 * 24 * 3600 = 103680 \ s$

The mass of the planet is $m_p = 9.38*10^{24} kg$

Generally the period of the moon is mathematically represented as

$T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }$

Here G is the gravitational constant with value

$G = 6.67 *10^{-11} \ N \cdot m^2/kg^2$

=> $T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }$

=> $103680 = 2 * 3.142 * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }$

=> $272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}$

=> $r = \sqrt[3]{ 1.7031241*10^{23}}$ j

=> $r = 55430496 \ m$

8 0

3 years ago

Xelga [282]

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

$\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}$

Where U is potential energy and K is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

$\displaystyle U_i = K_f$

Substitute and solve for final velocity:
$\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\ 2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ & =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}$

In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.

8 0

2 years ago