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4 years ago

Which of these most likely causes an annular solar eclipse?

Physics

2 answers:

Dmitrij [34]4 years ago

6 0

Answer:

C.the moon at apogee coming between Earth and the sun along a straight line

Explanation:

As we know that Earth rotate about the sun in elliptical orbit as one of the planet.

While moon rotate about the Earth similarly as one of the satellite around it

now when moon, Sun and Earth all lie in the same line such that position of moon is between the positions of Sun and Earth then in this case the light coming from sun is totally bounded by the position of moon

This situation of moon is known as Solar Eclipse

so here during the motion if the moon at apogee coming between Earth and the sun along a straight line then it is the most likely cause of annular solar eclipse.

sattari [20]4 years ago

5 0

C.the moon at apogee coming between Earth and the sun along a straight line

So, option C is your most accurate answer.

Hope this helps!

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irakobra [83]

Transverse waves are always characterized by particle motion being perpendicular to wave motion. A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction which the wave moves

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3 years ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Thepotemich [5.8K]

Answer:

$T=6.75s$

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on and the second when the rocket's engines is off. So, we need to know the height ( $h_1$ ) that the rocket reaches while its engine is on and we need to know the distance ( $h_2$ ) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

$h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\$

and the final speed is:

$v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T$

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

$v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2$

The sum of these heights will give us the total height, which is known:

$h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s$

This is the time that its needed in order for the rocket to reach the required altitude.

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3 years ago

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kompoz [17]

The answer is

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3 years ago

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4 years ago

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olga nikolaevna [1]

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3 years ago