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3 years ago

What change will always result in an increase in gravitational force between two objects

Physics

1 answer:

tekilochka [14]3 years ago

6 0

The gravitational force between two object depends on their masses and on their distance.

Since the formula is

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

If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.

So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.

So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase

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2. What is one reason why the author includes information about religious beliefs during the time of Johannes Fabricius' discove

Lesechka [4]

B. to show that there was conflict between what scientists were observing about the universe and what religion taught them

8 0

2 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

2 years ago

A man wishes to pull a crate 15m across a rough floor by exerting a force of 100 N. The

Darina [25.2K]

Answer:

option (E) is correct.

Explanation:

Work done is defined as the product of force and the distance in the direction of force.

force, f = 100 N

Coefficient of friction, = 0.25

distance = 15 m

So, net force F = f - friction force

F = 100 - 0.25 x m g

Work = (100 - 0.25 mg) x d cosθ

For minimum work, the angle should be maximum.

So, the value of θ is 76°.

thus, option (E) is correct.

3 0

2 years ago

What is the frequency of a wave that has a speed of 5 m/s and a wavelength of 0.5meter?

Debora [2.8K]

Explanation:

frequency =speed/wavelength

=5/0.5=10Hz

5 0

2 years ago

I am trying to find the magnitude of a resultant vector. Do i take inconsideration the negatives when i find the x & y compo

attashe74 [19]

Absolutely ! If you have two vectors with equal magnitudes and opposite
directions, then one of them is the negative of the other. Their correct
vector sum is zero, and that's exactly the magnitude of the resultant vector.

(Think of fifty football players pulling on each end of the rope in a tug-of-war.
Their forces are equal in magnitude but opposite in sign, and the flag that
hangs from the middle of the rope goes nowhere, because the resultant
force on it is zero.)

This gross, messy explanation is completely applicable when you're totaling up
the x-components or the y-components.

4 0

3 years ago