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2 years ago

Is the moon flat tell me if it it

Physics

2 answers:

QveST [7]2 years ago

6 0

I don’t think it is flat

yan [13]2 years ago

6 0

Answer: The moon is not flat.

Explanation: To the eye, the moon appears round, and it's natural to assume that it is actually spherical in shape with every point on its surface equidistant from its center like a big ball. The shape of the moon is that of an oblate spheroid, meaning it has the shape of a ball that is slightly flattened.

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An object at rest starts accelerating.

Shalnov [3]

Answer:

<u>We are given: </u>

initial velocity (u) = 0 m/s

final velocity (v) = 10 m/s

displacement (s) = 20 m

acceleration (a) = a m/s/s

<u>Solving for 'a'</u>

From the third equation of motion:

v² - u² = 2as

replacing the variables

(10)² - (0)² = 2(a)(20)

100 = 40a

a = 100 / 40

a = 2.5 m/s²

6 0

3 years ago

We can see Objects because of

Amiraneli [1.4K]

Answer:

I believe it's A.)

Explanation:

Although light comes into our atmosphere through refraction, it reaches our eyes only through reflection from objects. So when light rays reflect off an object and enter the eyes through the cornea you can then see that object.

Hope this helps you out : )

7 0

3 years ago

develop a plan to test the hardness of a sample of feldspar using the following items glass plate copper penny and streak plate

Artemon [7]

Stroke then 3883 you choke <3333333333

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2 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago

The three basic economic questions societies face are (1) what goods and services to produce, (2) how to produce those goods and

lord [1]

Answer:

For whom are goods and services to be produced? In other words, who gets what?

What should we produce?

For whom should we produce it?

Explanation:

8 0

2 years ago