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3 years ago

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How much energy Δ???? would be required to move the Moon from its present orbit around Earth to a location that is twice as far

away? Assume the Moon’s orbit around Earth is nearly circular and has a radius of 3.84×108 m. Δ

1 answer:

Zinaida [17]3 years ago

7 0

Answer:

$W = 1.9 \times 10^{28} J$

Explanation:

As we know that total energy of satellite is given as

$E = -\frac{GM_1M_2}{2r}$

now we know that work done to change the orbit is equal to change in the total energy of Earth - Moon system

so we know that

$W = E_f - E_i$

$W = -(\frac{GM_1M_2}{2r_2}) + (\frac{GM_1M_2}{2r_1})$

$W = \frac{GM_1M_2}{2}(\frac{1}{r_1} - \frac{1}{r_2})$

$W = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})(5.98 \times 10^{24})}{2}(\frac{1}{3.84 \times 10^8} - \frac{1}{7.68 \times 10^8})$

$W = 1.9 \times 10^{28} J$

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A woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 50.2 s.

lorasvet [3.4K]

Explanation:

(a) The period of a wave is the time required for one complete cycle. In this case, we have the time of five cycles. So:

$T=\frac{t}{n}\\\\T=\frac{50.2s}{5}\\T=10.04s$

(b) The frequency of a wave is inversely proportional to its period:

$f=\frac{1}{T}\\f=\frac{1}{10.04s}\\f=0.01Hz$

(c) The wavelength is the distance between two successive crests, so:

$\lambda=30.2m$

(d) The speed of a wave is defined as:

$v=f\lambda\\v=(0.1Hz)(30.2m)\\v=3.02\frac{m}{s}$

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3 years ago

When the moon is between the earth and the sun what moon phase will this be

Aleksandr [31]

Answer: The new moon phase occurs when the Moon is directly between the Earth and Sun. A solar eclipse can only happen at new moon. A waxing crescent moon is when the Moon looks like crescent and the crescent increases ("waxes") in size from one day to the next. This phase is usually only seen in the west.

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3 years ago

If v =5.00 meters/second and makes an angle of 60 with the negative direction of the y-axis coalculate the possible values of vx

Sever21 [200]

An angle of 60 degrees with the negative y-axis could mean 60 degrees clockwise or counterclockwise, which translates to two possible angles (starting from the positive x-axis and moving counterclockwise) of 210 degrees or 330 degrees.

Then the horizontal component $v_x$ of a velocity vector $\mathbf v$ with magnitude $5.00\,\dfrac{\mathrm m}{\mathrm s}$ could be one of two expressions:

$v_x=\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\cos210^\circ=-4.33\,\dfrac{\mathrm m}{\mathrm s}$

$v_x=\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\cos330^\circ=4.33\,\dfrac{\mathrm m}{\mathrm s}$

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3 years ago

Suppose a star the size of our Sun, but with mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 7.0 days.

kvasek [131]

Answer:

Its rotation will be 3.89x10⁴ rad/s.

Explanation:

We can find the rotation speed by conservation of the angular momentum:

$L_{i} = L_{f}$

$I_{i}\omega_{i} = I_{f}\omega_{f}$ (1)

The initial angular speed is:

$\omega_{i} = \frac{1 rev}{7 d} = 0.14 \frac{rev}{d}$

The moment of inertia (I) of a sphere is:

$I = \frac{2}{5}mr^{2}$ (2)

Where m is 9 times the sun's mass and r is the sun's radius

By entering equation (2) into (1) we have:

$\frac{2}{5}m_{i}r_{i}^{2}\omega_{i} = \frac{2}{5}m_{f}r_{f}^{2}\omega_{f}$

$9m_{sun}(696342 km)^{2}0.14\frac{rev}{d} = \frac{3}{4}9m_{sun}(13 km)^{2}\omega_{f}$

$\omega_{f} = \frac{4}{3}*0.14 \frac{rev}{d}(\frac{696342 km}{13 km})^{2} = 5.36 \cdot 10^{8} \frac{rev}{d}*\frac{1 d}{24 h}*\frac{1 h}{3600 s}*\frac{2\pi rad}{1 rev} = 3.89 \cdot 10^{4} rad/s$

Hence, its rotation will be 3.89x10⁴ rad/s.

I hope it helps you!

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4 years ago

Physics help pleaseee

Ilia_Sergeevich [38]

Answer:

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the second questions answer is- the atmospheric pressure increases.

Explanation:

I hope that this is what you were looking for.

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3 years ago