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2 years ago

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A rocket is fired in deep space, where gravity is negligible. If the rocket has a mass of 6000 kg and ejects gas at a relative v

elocity of 2000 m/s, how much gas must it eject in the first second to have an initial acceleration of 25.0 m/s?

1 answer:

shutvik [7]2 years ago

6 0

25 miles per second worth of gas

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Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air

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The frequency of a sound is whatever frequency leaves the source. It doesn't change.

Voiced of swimmers at the pool don't change frequency in or out of the water. Only their speed and wavelength change.

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If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g

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Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408 × 10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M_1, M_2$ are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

$\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}$

$\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2$

Since $M_2 = 2m_2$ and $r = R/3$

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2 years ago

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Answer: 9130 joules

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Workdone by wheelbarrow = ?

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Recall that power is the rate of doing work. Thus, power is workdone divided by time taken.

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Workdone = 830 watts x 11 seconds

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Thus, 9130 joules of work is required to get the wheelbarrow across the yard.

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Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly

loris [4]

Answer: $36.86\°$

Explanation:

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We need to find the angle $\theta$ at which Ferdinand jumps.

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Substituting (3) in (4):

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Clearing $\theta$ :

$\theta=cos^{-1} (\frac{4 m/s}{5 m/s})$

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