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3 years ago

What happens to the magnitude of the gravitational force as the distance between two bodies increase?

Physics

1 answer:

Anna [14]3 years ago

7 0

Answer:

The magnitude of the force will decrease

Explanation:

The gravitational force is one of the four fundamental forces of nature. It is an attractive force exerted between every object having mass.

Its magnitude is given by the equation:

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

m1 is the mass of the first object

m2 is the mass of the second object

r is the separation between the objects

As we see from the equation, the magnitude of the gravitational force is inversely proportional to the square of the distance between the objects:

$F\propto \frac{1}{r^2}$

Therefore, this means that as the distance between two bodies increases, the gravitational force will decrease.

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Additional Problem: A simple pendulum, consisting of a string (of negligible mass) of length L with a small mass m at the end, i

kykrilka [37]

Answer:

a) v = √ 2gL abd b) θ = 45º

Explanation:

a) for this part we use the law of conservation of energy,

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Em₀ = U = mg h

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Em₂ = ½ m v²

Em₀ = Em₂

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b) the definition of power is the relationship between work and time, but work is the product of force by displacement

P = W / t = F. d / t = F. v

If we use Newton's second law, with one axis of the tangential reference system to the trajectory and the other perpendicular, in the direction of the rope, the only force we have to break down is the weight

sin θ = Wt / W

Wt = W sin θ

This force is parallel to the movement and also to the speed, whereby the scalar product is reduced to the ordinary product

P = F v

The equation that describes the pendulum's motion is

θ = θ₀ cos (wt)

Let's replace

P = (W sin θ) θ₀ cos (wt)

P = W θ₀ sint θ cos (wt)

We use the equation of rotational kinematics

θ = wt

P = Wθ₀ sin θ cos θ

Let's use

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This expression is maximum when the sine has a value of one (sin 2θ = 1), which occurs for 90º,

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2 years ago

A 6.0 kg bowling ball moving at 3.5m/s to the right makes a collision, head-on, with a stationary 0.70 kg bowling pin. If the ba

devlian [24]

Answer:

The velocity of the pin will be 6.26 m/s in the right direction.

Explanation:

Let's use the momentum conservation equation.

$p_{i}=p_{f}$

Initially, we have:

$p_{i}=m_{b}*v_{ib}$

Where:

m(b) is the ball mass
v(ib) is the initial velocity of the ball

Now, the final momentum will be:

$p_{f}=m_{b}*v_{fb}+m_{p}*v_{fp}$

Where:

m(p) is the pin mass
v(fb) is the final velocity of the ball

v(fp) is the final velocity of the pin

Then, using the equation of the conservation we have:

$m_{b}*v_{ib}=m_{b}*v_{fb}+m_{p}*v_{fp}$

$6*3.5=6*2.77+0.7*v_{fp}$

$v_{fp}=6.26 m/s$

Therefore the velocity of the pin will be 6.26 m/s in the right direction.

I hope it helps you!

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3 years ago