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Leni [432]
2 years ago
12

In the figure , if Q = 30 uC , q = 5.0 uC , and d = 30 cm , what is the magnitude of the electrostatic force on g?

Physics
1 answer:
lana [24]2 years ago
3 0

Answer:

F =  k\frac{q_1q_1}{ {d}^{2} }  \\  = 9.0 \times  {10}^{9}  (N {m}^{2}  {C}^{ - 2}) \frac{30 \times  {10}^{ - 6} (C)5.0 \times {10}^{ - 6} (C)}{ {0.3}^{2} } N

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The force of friction occurs primarily because:
solniwko [45]

Answer:

B

Explanation:

Friction is a force that opposes motion between any surfaces that are touching. Friction occurs because no surface is perfectly smooth Friction produces heat because it causes the molecules on rubbing surfaces to move faster and have more energy.

3 0
2 years ago
WILL MARK BRAINLIEST! PLS ANSWER ASAP
Montano1993 [528]

Answer

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Explanation:

5 0
3 years ago
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A ball with a momentum of 16 kg.M/s strikes a ball at rest. What is the total momentum of both the balls after the collision.
melomori [17]

Answer:

Total momentum = 16 Kgm/s

Explanation:

Let the momentum of the two balls be A and B respectively.

Momentum A = 16 kgm/s

Momentum B = 0 kgm/s (since the ball is at rest).

Total momentum = A + B

Total momentum = 16 + 0

Total momentum = 16 Kgm/s

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

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8 0
2 years ago
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What two types of evidence are used to classify orgianisms?
jeyben [28]
One is their traits and their characterists that they have in common 
6 0
3 years ago
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A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.
NemiM [27]

Answer:

\dfrac{K_t}{K_r}=\dfrac{5}{2}

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, \omega=2.8\ rad/s

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

K_t=\dfrac{1}{2}mv^2

K_t=\dfrac{1}{2}m(r\omega)^2

K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2

The rotational kinetic energy of the ball is :

K_r=\dfrac{1}{2}I \omega^2

K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2

K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2

Ratio of translational to the rotational kinetic energy as :

\dfrac{K_t}{K_r}=\dfrac{5}{2}

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0
3 years ago
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