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2 years ago

6

A charge Q is transferred from an initially uncharged plastic ball to an identical ball 10.0 cm away. The force of attraction is

then 20.0 mN. Find magnitude of Q, give answer in C.

1 answer:

victus00 [196]2 years ago

5 0

Answer:

a charge Q is transferred from an initially uncharged

Explanation:

Hope this helps!

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Find the electric field at a point midway between two charges of +40.0 × 10−9 c and +60.0 × 10−9 c separated by a distance of 30

Anna [14]

The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
$E=k_e \frac{q}{r^2}$
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).

Let's calculate the electric field generated by the first charge:
$E_1 = (8.99 \cdot 10^9 Nm^2 C^{-2} ) \frac{+40.0 \cdot 10^{-9} C}{(0.15 m)^2}=1.6 \cdot 10^4 N/C$

While the electric field generated by the second charge is
$E_2 = (8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{+60.0 \cdot 10^{-9} C}{(0.15 m)^2}=2.4 \cdot 10^4 N/C$

Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
$E=E_2 - E_1 = 2.4 \cdot 10^4 N/C - 1.6 \cdot 10^4 N/C = 8000 N/C$

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Explanation:

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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

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2 years ago

Riders in a carnival ride stand with their backs against the wall of a circular room of diameter

Veseljchak [2.6K]

Answer:

option C

Explanation:

given,

diameter of circular room = 8 m

rotational velocity of the rider = 45 rev/min

= $45 \times \dfrac{2\pi}{60}$

=4.712 rad/s

here in this case normal force is equal to centripetal force

N = m r ω²

N = m x 4 x 4.712²

N = 88.83m

frictional force = μ N

= 88.83m x μ

now, for the body to not to slide

gravity force is equal to frictional force

m g = 88.83 m x μ

g = 88.83 x μ

9.8 = 88.83 x μ

μ = 0.11

hence, the correct answer is option C

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3 years ago