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2 years ago

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A dwarf planet has a mass of 0.0045 times that of the Earth and a diameter on average 0.19 times that of the Earth. What is the

escape velocity of the dwarf planet? (Type in the numerical answer and unit, e.g. 10m/s)

1 answer:

tatuchka [14]2 years ago

8 0

The escape velocity of the dwarf planet is 1,721.8 m/s.

The given parameters:

<em>Mass of the dwarf planet, m = 0.0045 M</em>
<em>Mass of the Earth = 5.98 x 10²⁴ kg</em>
<em>Diameter of the planet, d = 0.19 D</em>
<em>Diameter of the Earth, D = 12,742 km</em>

The mass of the of the dwarf planet is calculated as follows;

$m = 0.0045 \times 5.98 \times 10^{24} \ kg\\\\m = 2.69\times 10^{22} \ kg$

The radius of the dwarf planet is calculated as follows;

$r = \frac{0.19 D}{2} \\\\r = \frac{0.19 \times 12, 742, 000}{2} \\\\r = 1,210,490 \ m$

The escape velocity of the dwarf planet is calculated as follows;

$v _e = \sqrt{\frac{2GM}{r} } \\\\v_e = \sqrt{\frac{2\times 6.67 \times 10^{-11} \times 2.69 \times 10^{22}}{1,210,490}}\\\\v_e = 1,721.8 \ m/s$

Learn more about escape velocity here: brainly.com/question/13726115

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A super ball is dropped from a height of 100 feet. Each time it bounces, it rebounds half the distance it falls. How many feet w

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The total distance travelled by the ball after the fourth impact is 275 feet.

<u>Explanation:</u>

Given-

Height, h = 100 feet

Rebounds half the distance

Distance in feet for the fourth time, x = ?

For the first time, the distance travelled by the ball is, x = 100 feet

For the second time, it will bounce up to 50 feet and fall upto 50 feet( half of 100 feet)

So, the distance travelled after the second impact, x = 100 + 50 + 50 = 200 feet

For the third time, it will bounce up to 25 feet and fall upto 25 feet( half of 50 feet)

So, the distance travelled after the third impact, x = 200 + 25 + 25 = 250 feet

For the fourth time, it will bounce up to 12.5 feet and fall upto 12.5 feet( half of 25 feet)

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