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2 years ago

7

What event will produce the greatest increase in the gravitational force between the two masses?

1 answer:

TiliK225 [7]2 years ago

6 0

Answer:

Doubling the large mass

Explanation:

Doubling the destance bewteen the masses will simply make the gravitational force weaker

same with every answer exce

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What is the connection between electrons and protons

TEA [102]

Answer:

Electrons and protons are connected with one another in term of charge and size.

Explanation:

The size of electron and proton is always same in any atom but they possess opposite charge.
Electron in any atom carries negative charge where as proton carries the positive charge.
In any neutral atom the charge between electron and proton is balanced along with the size.
The nucleus of any atom bounds only proton and neutron but the electron is present revolving around the nucleus.

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3 years ago

1) La longitud del brazo de potencia de una palanca es de 0,0035 hectómetros y la del brazo de resistencia es de 55 centímetros.

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Answer:

Entonces seria 127 para vencer.

Explanation:

espero averte ayudado:-)

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3 years ago

12<br> Hot glass looks just like<br> ________ glass.<br> •Broken<br> •cold<br> •Shiny

Temka [501]

Answer:

shiny

Explanation:

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3 years ago

Read 2 more answers

If there is a break at any point in a series circuit the current will

Tomtit [17]

Answer:

not work

Explanation:

in a series circuit, everything meaning the electrons are flowing on one path, therefore, it wouldn continue to work.

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3 years ago

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul

GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that for spherical shell and pure rolling conditions

$v = R \omega$

$I = \frac{2}{3}mR^2$

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh$

$\frac{5}{6}mv^2 = mgh$

$h = \frac{5v^2}{6g}$

$h = \frac{5(8^2)}{6(9.81)} = 5.44 m$

Part b)

If ball is not rolling and just sliding over the hill then in that case

$\frac{1}{2}mv^2 = mgh$

$h = \frac{v^2}{2g}$

$h = \frac{8^2}{2(9.81)} = 3.16 m$

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3 years ago