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xeze [42]
3 years ago
12

A rocket of mass m is to be launched fromplanet X, which has a mass M and a radius R. What is the minimum speed that the rocket

must have for it to escape into space?
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:

The minimum speed = \sqrt{\frac{2GM}{R} }

Explanation:

The minimum speed that the rocket must have for it to escape into space is called its escape velocity. If the speed is not attained, the gravitational pull of the planet would pull down the rocket back to its surface. Thus the launch would not be successful.

The minimum speed can be determined by;

                      Escape velocity = \sqrt{\frac{2GM}{R} }

where: G is the universal gravitational constant, M is the mass of the planet X, and R is its radius.

If the appropriate values of the variables are substituted into the expression, the value of the minimum speed required can be determined.

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On a particle level, what happens when thermal conduction occurs within a solid?
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Faster particles bump into slower particles ( A )

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Read 2 more answers
A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can
GenaCL600 [577]

Answer:

0.02 m

Explanation:

R₁ = initial distance jumped by jumper = 7.4 m

R₂ = final distance jumped by jumper = ?

θ₁ = initial angle of jump = 45°

θ₂ = final angle of jump = 42.9°

v = speed at which jumper jumps at all time

initial distance jumped is given as

R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}

final distance jumped is given as

R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}

Dividing final distance by initial distance

\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}

\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}

R_{2} =7.38

distance lost is given as

d = R_{1} - R_{2}

d = 7.4 - 7.38

d = 0.02 m

8 0
3 years ago
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