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alexgriva [62]
3 years ago
13

Complete the statements about convection currents. Convection currents occur in fluids because of temperature and differences. C

onvection currents transfer energy throughout a fluid. They continue until all of the fluid is at the same temperature.
Physics
2 answers:
Sav [38]3 years ago
6 0

Answer

density; thermal

Explanation:

its right i just did it

marishachu [46]3 years ago
3 0

Answer:

density; thermal

Explanation:

Convection currents occur in fluids because of temperature and density differences. Convection currents transfer thermal energy throughout a fluid.

<em>When a fluid is heated, the molecules closest to the heat source gets heated first and their temperature is raised. The rise in temperature lowers their density and as such, they move to the upper portion of the fluid to be replaced by yet-to-be-heated molecules at the upper portion. </em>

<em>When the replacement molecules also get heated, the same process occurs and a to-and-from rotational movement of molecules, also known as convection current, is created within the fluid. The to and from movement, thus, transfer thermal energy within the fluid until all the molecules within the fluid are at the same temperature.</em>

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Ymorist [56]

Answer:

1.   <u>F = ma</u>  <em>F = 0.2kg * 20m/s² = 4Kg * m/s² =</em> 4N

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3.  <u>F = ma</u>  <em>F = 0.025Kg * 5m/s² =</em> 0.125N

4.  <u>F = ma</u>  <em>F = 50Kg * 4m/s² =</em> 200N

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8 0
2 years ago
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA
kkurt [141]

Answer:

The answer is "q=0.0945\,C".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}

Potential energy shifts:

= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\   =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\

=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5}  }{ 4,228 \times10^{5}} \right ) \\\\

=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\

\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0
3 years ago
Convert 25 cm to km using the method of dimensional analysis
pogonyaev

25 x 10^-5
= 0.00025

25 cm
= 0.00025 km
7 0
3 years ago
Read 2 more answers
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