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Sidana [21]
3 years ago
13

A small metal sphere has a mass of 0.11 g and a charge of -21.0 nC . It is 10 cm directly above an identical sphere with the sam

e charge. This lower sphere is fixed and cannot move.
Part A: What is the magnitude of the force between the spheres?
Part: B: If the upper sphere is released, it will begin to fall. What is the magnitude of its initial acceleration?
Physics
1 answer:
GrogVix [38]3 years ago
5 0

Answer:

A. the magnitude of the force between the spheres is 3.97 x 10⁻⁴ N

B. the magnitude of its initial acceleration is 5.83 m/s²

Explanation:

given information:

metal sphere's mass, m = 0.1 g = 1 x 10⁻⁴ kg

charge, q = -21 nC = -2.1 x 10⁻⁸

r = 10 cm = 0.1 m

What is the magnitude of the force between the spheres?

F₁₂ = k q₁q₂/r²

     = ( 9 x 10⁹) (-2.1 x 10⁻⁸)²/(0.1)²

     = 3.97 x 10⁻⁴ N

If the upper sphere is released, it will begin to fall. What is the magnitude of its initial acceleration?

mg - F₁₂ = ma

a = g - (F₁₂/m)

   = 9.8 - (3.97 x 10⁻⁴/1 x 10⁻⁴)

   = 5.83 m/s²

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The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
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The index of expansion then is 35/7.1 = 4.93

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Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

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Using law of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Along X- axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

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Along Y-axis

0=m_{1}v_{1}+m_{2}v_{2}

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v_{1}=-1.36j\ m/s

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(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

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(b). The direction of the first ball after the collision is 44.16° due south of east.

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