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3 years ago

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two 200 pound lead balls are separated by a distance 1m. both balls have the same positive charge q. what charge will produce an

electrostatic force.between the balls that is of the same order of magnitude as the weight of one ball?

1 answer:

Anna [14]3 years ago

5 0

Answer:

The charge is $q = 3.14 *10^{-4} \ C$

Explanation:

From the question we are told that

The mass of each ball is $m = 200 \ lb = \frac{200}{2.205} = 90.70 \ kg$

The distance of separation is $d = 1 \ m$

Generally the weight of the each ball is mathematically represented as

$W = m * g$

where g is the acceleration due to gravity with a value $g = 9.8 m/s^2$

substituting values

$W = 90.70 * 9.8$

$W = 889 \ N$

Generally the electrostatic force between this balls is mathematically represented as

$F_e = \frac{k * q_1* q_2 }{d^2}$

given that the the charges are equal we have

$q_1= q_2 = q$

So

$F_e = \frac{k * q^2 }{d^2}$

Now from the question we are told to find the charge when the weight of one ball is equal to the electrostatic force

So we have

$889 = \frac{9*10^9 * q^2}{1^2}$

=> $q = 3.14 *10^{-4} \ C$

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What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

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Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

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Where v is the velocity and r is the radius

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$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

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$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

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$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

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