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Arturiano [62]
3 years ago
14

Two equal point charges QQQ are separated by a distance ddd. One of the charges is released and moves away from the other due on

ly to the electrical force between them. When the moving charge is a distance 3ddd from the other charge, what is its kinetic energy?
Physics
1 answer:
lys-0071 [83]3 years ago
4 0

Answer:

The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

Explanation:

The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.

Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.

From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

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The gravitation force is 1/6 that of earth, and the ice in the same soft drink would float with 9/10 submerged option (B) is correct.

<h3>What is gravitational force?</h3>

All mass-bearing objects are attracted by gravitational force. Because it consistently attempts to bring masses together rather than push them apart, the gravitational force is referred to as attractive.

As we know, the gravitational force is given by:

\rm F = \dfrac{Gm_1m_2}{r^2}

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Weight of ice cube = weight of soft drink displaced

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Thus, the gravitation force is 1/6 that of earth, and the ice in the same soft drink would float with 9/10 submerged option (B) is correct.

An astronaut on Earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. If she were instead in a lunar module parked on the Moon where the gravitation force is 1/6 that of Earth, the ice in the same soft drink would float:A) with more than 9/10 submerged. B) with 9/10 submerged.C) with 6/10 submerged.D) totally submerged.

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1 year ago
What are the two factors that affect the friction force between two surfaces
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Answer:

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Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha
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The density of seawater at a depth where the pressure is 500 atm is 1124kg/m^3

Explanation:

The relationship between bulk modulus and pressure is the following:

B=\rho_0 \frac{\Delta p}{\Delta \rho}

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B is the bulk modulus

\rho_0 is the density at surface

\Delta p is the variation of pressure

\Delta \rho is the variation of density

In this problem, we have:

B=2.3\cdot 10^9 N/m^2 is the bulk modulus

\rho_0 =1100 kg/m^3

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Therefore, we can find the density of the water where the pressure is 500 atm as follows:

\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3

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