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Fofino [41]
2 years ago
11

The escape velocity at the surface of Earth is approximately 11 km/s. What is the mass, in units of ME (the mass of the Earth),

of a planet with three times the radius of Earth for which the escape speed is three times that for Earth?
Physics
1 answer:
SVETLANKA909090 [29]2 years ago
5 0

Answer:11 km/s

Explanation:

Given

Escape velocity at the surface of earth is 11 km/s

Escape velocity is given by

V_e=\sqrt{\frac{2GM}{R}}

Escape velocity at the surface of earth

11=\sqrt{\frac{2GM}{R}}--------------------1

If Escape velocity is three times and the radius is also the three times

V_e_2=\sqrt{\frac{2G(3M)}{3R}}

V_e_2=\sqrt{\frac{2GM}{R}}=V_e

i.e. V_e_2=11 km/s

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A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in
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Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a (Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

m - Mass of the box, measured in kilograms.

a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_{o}+a\cdot t), we expand the previous expression:

-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right) (Eq. 1b)

Where:

W - Weight, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,\frac{m}{s^{2}}, v_{o} = 1.37\,\frac{m}{s}, v = 0\,\frac{m}{s} and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

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f = 2.614\,N

The magnitude of the friction force exerted on the box is 2.614 newtons.

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