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Arlecino [84]
3 years ago
6

Pluto has a shape that is nearly round, and it orbits the Sun. It has five known moons. Why is it called a dwarf planet and not

a planet?
A.
It is too far from the Sun.
B.
Its orbit is not cleared of like-sized or larger objects.
C.
It is not large enough in size.
D.
It does not have rings.
Physics
1 answer:
Vikentia [17]3 years ago
6 0

Pluto has a shape that is nearly round, and it orbits the Sun. It has five known moons. It is called a dwarf planet and not a planet because its orbit is not cleared of like-sized or larger objects

Answer: Option B

<u>Explanation:</u>

According to the International Astronomical Union (IAU), Pluto has been mentioned as Dwarf Planet as it doesn't hold the standardisation to be counted as a Planet.

According to the standard criteria, a planet must have been surrounded by the orbit of the Sun, it must have significant mass to have a gravitational pull and hence, circular shape, and should have isolated surrounding.

But, Pluto has various celestial bodies in its vicinity that are larger in size. Hence, as concluded by the Union, any celestial body that does not has clear surrounding, is regarded as Dwarf Planet just like Pluto.

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C

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As your skateboard coasts uphill, your speed changes from 3 m/s to 1 m/s in
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a=-0.33\ m/s^2

Explanation:

<u>Accelerated Motion</u>

The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

\displaystyle a=\frac{\Delta v}{t}

Or, equivalently

\displaystyle a=\frac{v_f-v_o}{t}

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3 years ago
1. In a biology experiment the number of yeast cells is determined after 24 hours of growth at
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3 years ago
You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it
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(A) The speed just as it left the ground is 30.25 m/s

(B) The maximum height of the rock is 46.69 m

Explanation:

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speed of the rock at 14.8 m, u = 25 m/s

(a) Apply work energy theorem to find its speed just as it left the ground

work = Δ kinetic energy

F x d = ¹/₂mv² - ¹/₂mu²

mg x d = ¹/₂m(v² - u²)

g x d = ¹/₂(v² - u²)

gd = ¹/₂(v² - u²)

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v² = 2gd  + u²

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v² = 915.05

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B) Use the work-energy theorem to find its maximum height

the initial velocity of the rock = 30.25 m/s

at maximum height, the final velocity = 0

- mg x H = ¹/₂mv² - ¹/₂mu²

- mg x H = ¹/₂m(0) - ¹/₂mu²

- mg x H = - ¹/₂mu²

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H = u² / 2g

H = (30.25)² / 2(9.8)

H = 46.69 m

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3 years ago
Closed circuit
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