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zmey [24]
3 years ago
12

Which statements about the moon’s phases are true? Check all that apply.

Physics
2 answers:
patriot [66]3 years ago
7 0

Answer:

The same side of the moon always faces the earth

A new moon occurs when the Sun shines on the side of the moon not facing Earth.

A full moon happens when the Earth is between the Sun and the moon.

Explanation:

I just did the test

nikklg [1K]3 years ago
3 0

Answer:

A,C,F

1,3,6

Explanation:

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A 100 N force is applied to a 500 kg crate resting on frictionless wheels.
Alla [95]

Answer:

<h2>0.2 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{100}{500}  =  \frac{1}{5}  = 0.2 \\

We have the final answer as

<h3>0.2 m/s²</h3>

Hope this helps you

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The bulb

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3 years ago
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A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. Th
Karo-lina-s [1.5K]

Answer:

Work done by external force is given as

Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

Explanation:

As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car

so we will have

Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

now we have

W_{gravity} = -mg(Lsin\theta)

W_{friction} = -\mu mgcos(\theta) L

so from above equation

Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

so from above equation work done by external force is given as

Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

8 0
3 years ago
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