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Harrizon [31]
3 years ago
12

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

.78 μC is fired with an initial speed of 80.4 m/s directly toward the fixed charge. How far does the particle travel before its speed is zero?
Physics
1 answer:
Andreyy893 years ago
6 0

Answer:

d = 0.0306 m

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

U + K = constant

\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0

7.05 + 30.5 = \frac{0.266}{r}

r = 7.08 \times 10^{-3} m

so distance moved by the particle is given as

d = r_1 - r_2

d = 0.0377 - 0.00708

d = 0.0306 m

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2 years ago
How to do this question
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Answer:

(a) 10 m/s

(b) 22.4 m/s

Explanation:

(a) Draw a free body diagram of the car when it is at the top of the loop.  There are two forces: weight force mg pulling down, and normal force N pushing down.

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3 years ago
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The critical angle for a special type of glass in air is 30.8 ◦ . the index of refraction for water is 1.33. what is the critica
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When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted light, and all the light is reflected. The value of this angle is given by
\theta_c = \arcsin ( \frac{n_2}{n_1} )
where n2 and n1 are the refractive indices of the second and first medium, respectively.

In the first part of the problem, light moves from glass to air (n_a=1.00) and the critical angle is \theta_c = 30.8^{\circ}. This means that we can find the refractive index of glass by re-arranging the previous formula:
n_g=n_1 =  \frac{n_2}{\sin \theta_c}= \frac{1.00}{\sin 30.8^{\circ}}=1.95

Now the glass is put into water, whose refractive index is n_w = 1.33. If light moves from glass to water, the new critical angle will be
\theta_c = \arcsin ( \frac{n_2}{n_1} )=\arcsin( \frac{n_w}{n_g} )=\arcsin( \frac{1.33}{1.95} )=\arcsin(0.68)=43.0^{\circ}
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