If the Earth was the size of a basketball, about 24 centimeters in diameter, about how thick is the Earth's crust?

Which describes one event that causes an eclipse?

Burka [1]

Answer:

Earths shadow covering the moon would create a lunar eclipse.

Explanation:

because i just know

3 0

3 years ago

Suppose the electrons and protons in 1g of hydrogen could be separated and placed on the earth and the moon, respectively. Compa

MAXImum [283]

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

$F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}$

Where:

G= 6.674×10^−11 N · (m/kg)2

k = 8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

$\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}$ --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is 5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

$\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}$

Therefore

$\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}$

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

7 0

3 years ago

Power can be defined as

maria [59]

I think it’s D sorry If I’m wrong

4 0

3 years ago

Read 2 more answers

What is the difference between a learner’s license and an operator’s license?

Stells [14]

<span>A Learner’s license is available to those at least 15 years old that have passed the written and vision tests.
</span><span>An intermediate license, you must be 16 or 17 years old and you must have held a learner’s license for at least 12 months without receiving any traffic violations.</span>

5 0

3 years ago

Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500m3/sat a location 90 m above the lake sur

olga_2 [115]

Answer:

mechanical energy per unit mass is 887.4 J/kg

power generated is 443.7 MW

Explanation:

given data

average velocity = 3 m/s

rate = 500 m³/s

height h = 90 m

to find out

total mechanical energy and power generation potential

solution

we know that mechanical energy is sum of potential energy and kinetic energy

so

E = $\frac{1}{2}$ ×m×v² + m×g×h .............1

and energy per mass unit is

E/m = $\frac{1}{2}$ ×v² + g×h

put here value

E/m = $\frac{1}{2}$ ×3² + 9.81×90

E/m = 887.4 J/kg

so mechanical energy per unit mass is 887.4 J/kg

and

power generated is express as

power generated = energy per unit mass ×rate×density

power generated = 887.4× 500× 1000

power generated = 443700000

so power generated is 443.7 MW

3 0

3 years ago