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makvit [3.9K]
3 years ago
8

A rocket has landed on planet x, which has half the radius of earth. An astronaut onboard the rocket weighs twice as much on pla

net x as on earth. If the escape velocity for the rocket taking off from earth is u0 , then its escape velocity on planet x is
(a) 2u0
(b) 2u0
(c) u0
(d) u0 2
(e) u0 4
Physics
1 answer:
Nastasia [14]3 years ago
4 0

Answer:

Option (c) u0

Explanation:

The escape velocity has a formula as:

V = √(2gR)

Where V is the escape velocity,

g is the acceleration due to gravity

R is the radius of the earth.

Now, from the question, we were told that the escape velocity for the rocket taking off from earth is u0 i.e

V(earth) = u0

V(earth) = √(2gR)

u0 = √(2gR) => For the earth

Now, let us calculate the escape velocity for the rocket taking off from planet x. This is illustrated below below:

g(planet x) = 2g (earth) => since the weight of the astronaut is twice as much on planet x as on earth

R(planet x) = 1/2 R(earth) => planet x has half the radius of earth

V(planet x) =?

Applying the formula V = √(2gR), the escape velocity on planet x is obtained as follow:

V(planet x) = √(2g(x) x R(x))

V(planet x) = √(2 x 2g x 1/2R)

V(planet x) = √(2 x g x R)

V(planet x) = √(2gR)

The expression obtained for the escape velocity on planet x i.e V(planet x) = √(2gR), is exactly the same as that obtained for the earth i.e V(earth) = √(2gR)

Therefore,

V(planet x) = V(earth) = √(2gR)

But from the question, V(earth) is u0

Therefore,

V(planet x) = V(earth) = √(2gR) = u0

So, the escape velocity on planet x is u0

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Answer:

2.85 rad/s

Explanation:

5 cm = 0.05 m

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When dropping the 2nd object at a distance of 0.05 m from the center of mass, its corrected moments of inertia is:

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3 years ago
Why caring a body and moving with it is not a work done<br><br> Why :​
Alecsey [184]

Answer:

Work done on an object is equal to

FDcos(angle).

So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.

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Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction.

Explanation:

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A 20-kg block slides down a fixed rough curved track The block has a speed of 5 0 m/s after its height above a horizontal surfac
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U = 102.8 J (100 J to two significant digits)

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Answer:

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