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3 years ago

When the distance between two interacting objects doubles, the gravitational force is

Physics

1 answer:

Umnica [9.8K]3 years ago

5 0

The gravitational force will be one quarter.

The gravitational force between two objects is given by the formula

F=GMm/r^2

here, r is the distance between the objects.

Thus the gravitational force is inversely proportional to the square of the distance between the objects, Therefore if the distance between two objects is doubled the force will be one quarter.

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What happens to the velocity of a sound wave in air if the temperature of the air increases?

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The answer is D :) hope this helps

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3 years ago

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The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

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According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

PART B) Replacing the values given as,

$h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

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$v_f = 1.8486m/s$

Therefore the speed of the masses would be 1.8486m/s

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3 years ago

Suppose a proton ( = 1. 67×10^−27 kg) is confined to a box of width = 1. 00×10^−14 m (a typical nuclear radius).

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Answer:

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Explanation:

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3 years ago

If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he

forsale [732]

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q= mC_s \Delta T$
where m is the mass of the substance, Cs is its specific heat capacity and $\Delta T$ is the increase of temperature.

If we re-arrange the formula, we get
$C_s = \frac{Q}{m \Delta T}$
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
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3 years ago

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Si la fuerza de repulsión entre dos cargas es 18 × 1013

sattari [20]

Answer:

Explanation:

F = kQq/r²

r = √(kQq/F)

a) r = √(8.899(10⁹)(8)(4) / 18(10¹³)) = 0.0397749... m

r = 40 mm

b) r = √(8.899(10⁹)(12)(3) / 18(10¹³)) = 0.0421876... m

r = 42 mm

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3 years ago