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3 years ago

In the diagram, the Sun, Earth, and Moon are in perfect alignment. Which two conclusions can be drawn based on the diagram?

Physics

2 answers:

mihalych1998 [28]3 years ago

8 0

Answer:An observer on the Moon sees a solar eclipse. and An observer on Earth sees a lunar eclipse.

Explanation:from the moon the sun is completly if not just mostly invisible(solar eclipse) and from earth the moon would be in lunar eclipse from the earth blocking the sun.

AysviL [449]3 years ago

4 0

Answer:

I know one answer is an observer on earth sees a lunar eclipse

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If the Earth rotated more slowly about its axis, your apparent weight would

disa [49]

<span>If the Earth rotated more slowly about its axis, your apparent weight would
A) increase.
B) decrease.
C) stay the same.
D) be zero.

</span>A) increase.

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3 years ago

3 light-years =________.

KATRIN_1 [288]

   ( 3 yr) · (186,282.397 mile/s) · (86,400 s/day) · (365 day/yr)

   = (3 · 186,282.397 · 86,400 · 365) mile

   = 1.762380502 x 10¹³ miles

   = 1.8 x 10¹³ miles (rounded to the nearest trillion miles)

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3 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago

Fear me or i am glory i give you free p oints <br> \ /<br> _| |_

8_murik_8 [283]

Answer:

kayyy

Explanation:

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3 years ago

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Which is an example of an unstructured activity that promotes resistance training?

aleksandrvk [35]

One possible unstructured activity that promotes resistance training would be climbing playground equimpent - A.

This is by nature a unstructured ctivity. Furthermore, it promotes resistance training because you're forced to move and pull and push yourself.

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3 years ago