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2 years ago

13

The force of friction between an object and the surface upon which it is sliding is 14N and the coefficient of friction between

them is 0.72. What is the weight of the object?

1 answer:

Nat2105 [25]2 years ago

5 0

Answer:

14 x 0.27 = 3.78 is your answer

Explanation:

the question is asking for the weight of the object so you multiply and get 3.79

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The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the

seraphim [82]

Answer:

$x=\dfrac{D}{1+\sqrt{\dfrac{m}{M}}}$

Explanation:

Mass of moon = m

Mass of planet =M

We know that gravitational force given as

$F=G\dfrac{m_1m_2}{d^2}$

$F'=G\dfrac{m'M}{x^2}$

$F=G\dfrac{m'm}{(D-x)^2}$

Given that force is zero so

F=F'

$G\dfrac{m'm}{(D-x)^2}=G\dfrac{m'M}{x^2}$

$\dfrac{m}{(D-x)^2}=\dfrac{M}{x^2}$

$\dfrac{x}{D-x}=\sqrt{\dfrac{M}{m}}$

$\dfrac{D-x}{x}=\sqrt{\dfrac{m}{M}}$

$x=\dfrac{D}{1+\sqrt{\dfrac{m}{M}}}$

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2 years ago

Convert 5.7 miles to km

gayaneshka [121]

5.7 kilometers is equal to 3.5418157957528034 miles

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2 years ago

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A straight trail with a uniform inclination of 15 degrees leads from a lodge at an elevation of 600 feet to a mountain lake at a

9966 [12]

Answer:

The length of the trail = 22796 ft

Explanation:

From the ΔABC

AC = length of the trail = x

AB = 6100 - 600 = 5500 ft

Angle of inclination $\theta$ = 15°

$\sin \theta = \frac{AB}{AC}$

$\sin 15 = \frac{5900}{x}$

$x = \frac{5900}{0.2588}$

x = 22796 ft

Since x = AC = Length of the trail.

Therefore the length of the trail = 22796 ft

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2 years ago

A 2-kg ball is thrown at 3 m/s. What is the ball's momentum? *

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Answer:

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mass (m) =2kg

velocity (v) =3m/s

momentum (p) =?

Form

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the momentum of ball's =6kg.m/s

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2 years ago

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