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Afina-wow [57]
3 years ago
10

A boy exerts a 36-N horizontal force as he pulls a 52-N sled across a cement sidewalk at a constant speed. What is the coefficie

nt of kinetic friction between the sidewalk and the metal sled runners? Ignore air resistance.
Physics
2 answers:
Vesna [10]3 years ago
7 0

Answer:

0.69

Explanation:

Sum of the forces on the sled in the y direction:

∑F = ma

Fn - 52 N = 0

Fn = 52 N

Sum of the forces on the sled in the x direction:

∑F = ma

36 N - Fn μ = 0

Fn μ = 36 N

μ = 36 N / Fn

Substitute:

μ = 36 N / 52 N

μ = 0.69

kupik [55]3 years ago
6 0

Answer:

cofficient of kinetic friction is 0.6923

Explanation:

Given Data

Coefficient of kinetic friction  μk = f_{k}/N where fk is force of kinetic friction and N is normal reaction;

we have N = 52 N, and fk = 36 N ;

                μk = f_{k}/N

                      = 36/52

                      = 0.6923

    cofficient of kinetic friction is 0.6923.

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During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

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