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Sidana [21]
3 years ago
15

An electric field of 280000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -

7.9C at this spot?
Physics
1 answer:
nordsb [41]3 years ago
4 0
<h2>The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.</h2>

Explanation:

Electric field is the ratio of force and charge.

Electric field, E = 280000 N/C

Charge, q = -7.9 C

We have

                 E=\frac{F}{q}\\\\280000=\frac{F}{7.9}\\\\F=280000\times 7.9\\\\F=2.21\times 10^6N

The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.

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An object with a mass m slides down a rough 370 inclined plane where the coefficient of kinetic friction is 0.20. If the plane i
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Answer:

v \approx 9.312\,\frac{m}{s}

Explanation:

The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:

K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}

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\frac{1}{2}\cdot v^{2} = g\cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)

v = \sqrt{2\cdot g \cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}

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To demonstrate the tremendous acceleration of a top fuel dragracer, you attempt to run your car into the back of a dragster that
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a. 2v₀/a   b. 2v₀/a  

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a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.

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= 2v₀/a  

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