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miss Akunina [59]
3 years ago
8

If the sun were to vanish instantly would earth immediately fly out of its orbit explain why or why not

Physics
1 answer:
Zanzabum3 years ago
7 0
<span>Answer: No, because Einstein demonstrated that nothing can exceed the speed of light in a vacuum and for something to happen instantly over that distance would require that speed to be exceeded. If somehow the sun were to vanish, without explosive effects, an enormous gravity wave would begin travelling outward affecting the planets at the speed of light - thus taking about 8 minutes to reach earth. But that is irrelevant because the only way to remove all that matter would be total conversion of the mass to energy and that energy would totally destroy everything - after the same 8 minutes. Mike1942f · 9 years ago</span>
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Which of the following is not a process that forms metallic ore
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I believe that this question has the following choices to choose from:

placer deposits

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<span>fossil compaction                (answer)</span>

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4 years ago
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A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t
mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 \frac{m}{s^{2}}

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

K=U

\frac{mv^2}{2}=\frac{kx_{max}^2}{2}

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=

v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}

part 3

Acceleration can be find using Newton's second law:

F=ma

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

-kx=ma

a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s

So between 0 and 4.5 cm we have half a period:

t=\frac{T}{2}=0.66s

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3 years ago
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