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4 years ago

A 75.0 kg students sits 1.15 m away from a 68.4 kg student. What is the force of gravitational attraction between them?

Physics

1 answer:

AVprozaik [17]4 years ago

7 0

Answer:

F=2.589×10⁻⁷ Newtons

Explanation:

The gravitational force of attraction between two bodies is given by

F= Gm₁m₂/r²

Where m₁ and m₂ are the masses of the two bodies, G is the universal gravitational constant while r is the distance between the two bodies.

G=6.67408 × 10⁻¹¹m³kg⁻¹s⁻²

r=1.15m

m₁=75.0kg

m₂=68.4kg

Therefore F= (6.67408×10⁻¹¹m³kg⁻¹s⁻²×75.0kg×68.4kg)/(1.15m)²

F=2.589×10⁻⁷ Newtons

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= $\frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} } = \frac{106x}{x+53}$

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3 years ago

I need help with part D

natka813 [3]

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3 years ago

A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on

Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

$F_f = ma$ (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

$m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg$

And we can fidn the acceleration by using the formula:

$a=\frac{v-u}{t}$

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

$a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2$

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

$F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N$

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3 years ago

The displacement of a car is a function of time as follows: x(t)=25+3.0t², with x is in meters. Find the average velocity betwee

aniked [119]

Answer: 15m/s

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3 years ago

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4 years ago