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3 years ago

7

The Earth’s internal __________ source provides the energy for our dynamic planet, providing it with the driving force for on-go

ing disastrous events such as earthquakes and volcanic eruptions and for plate-tectonic motion.

1 answer:

Lesechka [4]3 years ago

4 0

The internal (heat) source

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What do scientists use to determine the temperature of a star?

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Read 2 more answers

Would you be more concerned if something you consider precious went through a

Dahasolnce [82]

Answer:

Explanation:

A chemical change.

Usually those are irreversible. Or they may be reversible, but the form they take may leave your object not the same as they started out.

A physical change might be just as deadly. If the object melted like a chocolate Easter Bunny then the object would be irreversible as well. Take a better example.

Suppose you are talking about a Gold Coin. If you heated it so it melted, the gold would retain its value, but the fact that it is a coin and valuable as such, means that it has lost that part of its value.

I really don't know. My instincts tell me that the chemical change is more dangerous, but I can't rule out the other choice..

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2 years ago

two astronauts are taking a spacewalk outside the International Space Station the first astronaut has a mass of 64 kg the second

Fittoniya [83]

Answer:

Approximately $0.88\; {\rm m \cdot s^{-1}}$ to the right (assuming that both astronauts were originally stationary.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Since momentum of this system (of the astronauts) conserved:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\end{aligned}$ .

Assuming that both astronauts were originally stationary. The total initial momentum of the two astronauts would be $0$ since the velocity of both astronauts was $0\!$ .

Therefore:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

The final momentum of the first astronaut ( $m = 64\; {\rm kg}$ , $v = 0.8\; {\rm m\cdot s^{-1}}$ to the left) would be $p_{1} = m\, v = 64\; {\rm kg} \times 0.8\; {\rm m\cdot s^{-1}} = 51.2\; {\rm kg \cdot m \cdot s^{-1}}$ to the left.

Let $p_{2}$ denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be $(p_{1} + p_{2})$ .

$\begin{aligned} & p_{1} + p_{2} \\ &= (\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

Hence, $p_{2} = (-p_{1})$ . In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude ( $51.2\; {\rm kg \cdot m \cdot s^{-1}}$ ) but opposite in direction (to the right versus to the left.)

Rearrange the equation $p = m\, v$ to obtain an expression for velocity in terms of momentum and mass: $v = (p / m)$ .

$\begin{aligned}v &= \frac{p}{m} \\ &= \frac{51.2\; {\rm kg \cdot m \cdot s^{-1}}}{64\; {\rm kg}} && \genfrac{}{}{0}{}{(\text{to the right})}{} \\ &\approx 0.88\; {\rm m\cdot s^{-1}} && (\text{to the right})\end{aligned}$ .

Hence, the velocity of the astronaut in question ( $m = 58.2\; {\rm kg}$ ) would be $0.88\; {\rm m \cdot s^{-1}}$ to the right.

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2 years ago