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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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padilas [110]

Answer:

Winner wins by 0.969 s

Explanation:

For the Porche:

Given:

Displacement of Porsche s = 400 m

Acceleration of Porsche a = 3.4 m/s^2

From Newton's second equation of motion,

s = ut + (1/2) a t^2 (u = 0 as the car was initially at rest)

Substituting the values into the equation, we have

t^2 = (2 * 400) / 3.4

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t = 15.33 s

For the Honda:

Displacement of Honda = 310 m

Acceleration of Honda = 3 m/s^2

Applying Newton's second equation of motion

s = ut + (1/2) a t^2 (u = 0 for same reason)

Substituting the values into the equation, we obtain

t^2 = (2 * 310) / 3

= 620 / 3

t = 14.37 s

Hence

The winner (honda) wins by a time interval of = 15.33 - 14.37    

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