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mamaluj [8]
3 years ago
5

How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (grav

itational acceleration on the Moon is about 1/6 of g on Earth)?
Physics
1 answer:
JulsSmile [24]3 years ago
8 0

Answer:

maximum height on moon is 6 times more than the maximum height on Earth

Explanation:

Let the Astronaut has its maximum speed by which he can jump is "v"

now for the maximum height that it can jump is given as

v_f^2 - v_i^2 = 2 aH

now from above equation we will have

0 - v^2 = 2(-g)H

now we have

H = \frac{v^2}{2g}

now if Astronaut jump on the surface of moon with same speed

then we know that the acceleration of gravity on surface of moon is 1/6 times the gravity on earth

so at surface of moon we have

0 - v^2 = 2(-g/6) H

now we have

H = \frac{6v^2}{2g}

so maximum height on moon is 6 times more than the maximum height on Earth

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A rock resting high in a cliff is an example of an object with what type of energy?
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Which of the following actions will keep the gravitational force between two objects unchanged?
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increasing the temperature of the object

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A city bus travels along its route from Jackson Street 8
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A huge cannon is assembled on an airless planet (ignore any effects due to the planet's rotation). The planet has a radius of 5.
Ganezh [65]

Answer:

The projectile's speed as it passes the satellite is 1497.8 m/s.

Explanation:

Given that,

Radius of planet r=5.00\times10^{6}\ m

Mass of planet m=3.95\times10^{23}\ kg

Speed = 2000 m/s

Height = 1000 km

We need to calculate the projectile's speed as it passes the satellite

Using conservation of energy

E_{1}=E_{2}

\dfrac{1}{2}mv_{1}^2+\dfrac{GmM}{r_{1}}=\dfrac{1}{2}mv_{2}^2+\dfrac{GmM}{R+h}

\dfrac{v_{1}^2}{2}+\dfrac{GM}{r_{1}}=\dfrac{v_{2}^2}{2}+\dfrac{GM}{R+h}

-\dfrac{v_{2}^2}{2}=-(\dfrac{GM}{R}-\dfrac{GM}{R+h}-\dfrac{v_{1}^2}{2})

v_{2}^2=v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})

v_{2}=\sqrt{v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})}

Put the value into the formula

v_{2}=\sqrt{2000^2+2\times6.67\times10^{-11}\times3.95\times10^{23}(\dfrac{1}{5.00\times10^{6}+1000\times10^{3}}-\dfrac{1}{5.00\times10^{6}})}

v_{2}=1497.8\ m/s

Hence, The projectile's speed as it passes the satellite is 1497.8 m/s.

4 0
3 years ago
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