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Aliun [14]
3 years ago
10

Which of the following lists the composition of the Earth’s outer core?

Physics
1 answer:
iris [78.8K]3 years ago
5 0
Well,

The outer core of the Earth is mostly composed of iron and nickel.

The correct option is C.
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<h2 /><h2>A) The ball will speed up.</h2>
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A 12 volt car battery is connected to a 3 ohm brake light. What is the current carrying energy to the lights?
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4 A

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The pressure exerted by 15m of liquid is 1500pa.The acceleration due to gravity g=10m/s^2.Calculate the density liquid.​
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Answer:

1500 divided by 150(15m x 10m/s^2) = 10

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A ball is dropped from a height of 12 feet and returns to a height that is one-half of the height from which it fell. How far wi
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The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t
Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

PE=\frac{GMm}{r}

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

m = \rho V

m = (1030Kg/m^3)(7*10^8m^3)

m = 7.210*10^{11}Kg

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m

PART A) Potential energy when the ocean is at its furthest point to the moon,

PE_1 = \frac{GMm}{r_1}

PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}

PE_1 = 9.05*10^{15}J

PART B) Potential energy when the ocean is at its closest point to the moon

PE_2 = \frac{GMm}{r_2}

PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}

PE_2 = 9.361*10^{15}J

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

\Delta KE = \Delta PE

\frac{1}{2}mv^2 = PE_2-PE_1

v=\sqrt{2(PE_2-PE_1)/m}

v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}

v = 29.4m/s

8 0
3 years ago
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