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creativ13 [48]
3 years ago
10

In complete sentences, describe what causes the phases of the moon. (4 points)

Physics
1 answer:
g100num [7]3 years ago
7 0
The light from the sun is reflected by the moon, we see the part of the moon that the sun shines on.The change in the phases are created from the rotation of the moon around earth.We see a full moon when the sun is opposite of the moon. The side the sun shines on the moon is what we see.
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Clara made a chart to summarize some of the evidence that supports the big bang theory.
cluponka [151]

Answer:

The temperature of the CMB is cooler, not hotter, than at the time of the big bang.

Explanation:

Hope it help

Mark me as brainliest

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2 years ago
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With a bit of algebraic reasoning find your gravitational acceleration toward any planet of mass M a distance d from its center.
grandymaker [24]

The acceleration due to gravity is given as:

                             g = GM/r²

<h3>Derivation of gravitational acceleration:</h3>

According to Newton's second law of motion,

F = ma

where,

F = force

m = mass

a = acceleration

According to Newton's law of gravity,

F<em>g </em>= GMm/(r + h)²

F<em>g = </em>gravitational force

From Newton's second law of motion,

F<em>g </em>= ma

a = F<em>g</em>/m

We can refer to "a" as "g"

a = g = GMm/(m)(r + h)²

g = GM/(r + h)²

When the object is on or close to the surface, the value of g is constant and height has no considerable impact. Hence, it can be written as,

g = GM/r²

Learn more about gravitational acceleration here:

brainly.com/question/2142879

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5 0
2 years ago
What is the direction of the sum of these two vectors? ​
gayaneshka [121]

Answer:

what is the direction of the sum of these two vectors?

7 0
3 years ago
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A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

\sum \tau = F*d

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

F*(27-6)= 6*600

F = \frac{6*600}{21}

F= 171.42 lb

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0
3 years ago
An object of mass 30 kg is in free fall in a vacuum where there is no air resistance. Determine the acceleration of the object.
velikii [3]

Answer:

In free fall, mass is not relevant and there's no air resistance, so the acceleration the object is experimenting will be equal to the gravity exerted. If the object is falling on our planet, the value of gravity is approximately 9.81ms2 .

7 0
3 years ago
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