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3 years ago

11

The amount of the lighted side of the moon you can see is the same during

1 answer:

attashe74 [19]3 years ago

5 0

<u>Answer:</u>

The amount of the lighted side of the moon you can see is the same during "how much of the sunlit side of the moon faces Earth".

<u>Explanation:</u>

The Moon is in sequential rotation with Earth, and thus displays the Sun, the close side, always on the same side. Thanks to libration, Earth can display slightly greater than half (nearly 59 per cent) of the entire lunar surface.

The side of the Moon facing Earth is considered the near side, and the far side is called the reverse. The far side is often referred to as the "dark side" inaccurately but it is actually highlighted as often as the near side: once every 29.5 Earth days. During the New Moon the near side becomes blurred.

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Mathphys Help help help

Keith_Richards [23]

Answer:

1030 mph

Explanation:

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First, in the x (east) direction:

vₓ = 335 mph + 711 cos 19° mph

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And in the y (north) direction:

vᵧ = 0 mph + 711 sin 19° mph

vᵧ = 231 mph

The net speed can be found with Pythagorean theorem:

v² = vₓ² + vᵧ²

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3 years ago

A hiker climbs a mountain. Starting at the base of the mountain, he first moved up 520m at a 32.0 degree angle. What is the fina

balu736 [363]

Answer:

$\displaystyle \vec{d}=$

Explanation:

<u>Displacement Vector</u>

Suppose an object is located at a position

$\displaystyle P_1(x_1,y_1)$

and then moves at another position at

$\displaystyle P_2(x_2,y_2)$

The displacement vector is directed from the first to the second position and can be found as

$\displaystyle \vec{d}=$

If the position is given as magnitude-angle data ( z , α), we can compute its rectangular components as

$\displaystyle x=z\ cos\alpha$

$\displaystyle y=z\ sin\alpha$

The question describes the situation where the initial point is the base of the mountain, where both components are zero

$\displaystyle P_1(0,0)$

The final point is given as a 520 m distance and a 32-degree angle, so

$\displaystyle x_2=520\ cos32^o= 440.99\ m$

$\displaystyle y_2=520\ sin32^o=275.6\ m$

The displacement is

$\displaystyle \vec{d}=$

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Answer:

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From what height is the ball originally dropped is h= 913.90 (m).

Explanation:

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