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NARA [144]
3 years ago
8

During a 0.001 s interval while it is between the plates, the change of the momentum of the electron Δp with arrow is < 0, -8

.80e-17, 0 > kg m/s. What is the electric field between the plates?
Physics
1 answer:
Delvig [45]3 years ago
8 0

Answer:

5.5 x 10^5 N/C

Explanation:

t = 0.001 s

Δp = - 8.8 x 10^-17 kg m /s

Force is equal to the rate of change of momentum.

F = Δp / Δt

F = (8.8 x 10^-17) / 0.001 = 8.8 x 10^-14 N

q = 1.6 x 10^-19 C

Electric field, E = F / q = (8.8 x 10^-14) / (1.6 x 10^-19)

E = 5.5 x 10^5 N/C

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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
Andreyy89

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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A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

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Part c)

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Part d)

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3 years ago
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