Earth's core is the source of the energy that drives the movement of tectonic

Physics

2 answers:

Ostrovityanka [42]2 years ago

5 0

Answer:

B. Convection

D. Conduction

Explanation:

Conduction and convection are the two most prominent processes that helps transfer energy outward to the earth's crust.

Energy within the core is a function of the radioactive decay and frictional heating.
Also, heat that accreted during the formation of the earth is a significant source of internal energy.
The heat is conducted away by the process of convection. This is possible due to temperature differences between different parts of the earth
Conduction is made made possible due to the metallic bodies in the core and other part of the inner earth.

dezoksy [38]2 years ago

4 0

Answer:

B and D

Explanation:

I just did it on a p e x. In case your not sure

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The orbital radius of the Earth (from Earth to Sun) is 1.496 x 10^11 m.

mrs_skeptik [129]

Explanation:

The orbital radius of the Earth is $r_1=1.496\times 10^{11}\ m$

The orbital radius of the Mercury is $r_2=5.79 \times 10^{10}\ m$

The orbital radius of the Pluto is $r_3=5.91 \times 10^{12}\ m$

We need to find the time required for light to travel from the Sun to each of the three planets.

(a) For Sun -Earth,

Kepler's third law :

$T_1^2=\dfrac{4\pi ^2}{GM}r_1^3$

M is mass of sun, $M=1.989\times 10^{30}\ kg$

So,

$T_1^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 1.496\times 10^{11}\\\\T_1=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times1.496\times10^{11}}\\\\T_1=2\times 10^{-4}\ s$

(b) For Sun -Mercury,

$T_2^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.79 \times 10^{10}\ m\\\\T_2=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.79 \times 10^{10}}\ m\\\\T_2=1.31\times 10^{-4}\ s$

$T_3^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.91 \times 10^{12}\\\\T_3=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.91 \times 10^{12}}\\\\T_3=1.32\times 10^{-3}\ s$

8 0

3 years ago

A ball is dropped from a height of 10m. At the same time, another ball is thrown

soldi70 [24.7K]

5.1 m

Explanation:

Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by

$y_1 = 10 - \frac{1}{2}gt^2$ (1)

where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground $y_2,$ is given by