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nasty-shy [4]
3 years ago
11

A mad scientist wants to collect massive amounts of charge on basketball sized aluminum balls. The scientist wants to place 6 C

on one ball and 14 C on another. However, the maximum force between the balls must be limited to 8000 N. At what distance must the balls be placed apart?
____________ m
Physics
1 answer:
vitfil [10]3 years ago
4 0

The distance between two basket ball sized aluminium balls is 9714 m.

Explanation:

Coulomb's law, or Coulomb's inverse-square law, is an experimental law of physics that quantifies the amount of force between two stationary, electrically charged particles. The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force .

Coulomb's law formula => F = (k * Qb1 * Qb2)/r²

Given data :-

charge on ball 1 Qb1 = 6C

charge on ball 2 Qb2 = 14C

Force exerted F = 8000 N

k =  8.988 x 10^9 Nm²C−²(coulomb's constant).

substituting given values in the coulomb's formula

8000 = (( 8.988 x 10^9)*6*14)/r²

shifting r and 8000 to other sides

r² = (756 * 10^9)/8000

r = 9714 m.

Therefore the distance between two balls is r = 9714 m.

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The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
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Answer:

 k = 11,564 N / m,   w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

 let's apply the equilibrium condition at this point

                 

Axis y

          W_{y} - Fr = 0

          Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

             sin 46 = W_{y} / W

             W_{y} = W sin 46

     

 we substitute

           mg sin 46 = k y

           k = mg / y sin 46

If the length of the bar is L

          sin 46 = y / L

           y = L sin46

 

we substitute

           k = mg / L sin 46 sin 46

           k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

           k = 1.18 9.8 / 1

           k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

          Em₀ = U = mgy

end point. Point at 30º

         Em_{f} = K -Ke = ½ I w² - ½ k y²

          em₀ = Em_{f}

          mgy = ½ I w² - ½ k y²

          w = √ (mgy + ½ ky²) 2 / I

the height by 30º

           sin 30 = y / L

           y = L sin 30

           y = 0.5 m

the moment of inertia of a bar that rotates at one end is

          I = ⅓ mL 2

          I = ½ 1.18 12

          I = 0.3933 kg m²

let's calculate

          w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

          w = 6.06 rad / s

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