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4 years ago

Where does the heat come from that creates the motion for continental drift?

Physics

1 answer:

poizon [28]4 years ago

4 0

Answer:

The heat doesn't come from any of those the heat comes from the Sun

Explanation:

But the thing that holds the heat from the Sun is the Earth's atmosphere.

When he gets hot the carbon dioxide holds in the heat but when the when oxygen it meets the heat in the air it makes it become cooler

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The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits that are in a stead

Fed [463]

Answer:

a) Current.

Explanation:

The "junction rule" (Kirchhoff's first law) says that the value of the electric current entering any point in the circuit must be the same as the one leaving it. This comes as a consequence of the conservation of charge, the electric charge that comes in must come out (when speaking about currents, per unit time).

3 0

3 years ago

Which factors are used to calculate the kinetic energy of an object? Check all that apply.

Verizon [17]

Gravity and velocity

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3 years ago

Is a copper wire a conductor or insulator?

irina [24]

Answer:

its a conductor because it conducts electricity

Explanation:

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3 years ago

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Why do the planets move in fixed orbits around the Sun? A) The magnetic pull of the Sun holds them in their orbits. B) The gravi

SpyIntel [72]

Answer:its B

Explanation:

The suns gravity keeps the planets in their gravitational orbit.

7 0

3 years ago

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The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$

Therefore; the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

5 0

4 years ago