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3 years ago

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A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 22.0 m/s and returns the shot with the ball

traveling horizontally at 35.5 m/s in the opposite direction. (assume the initial direction of the ball is in the −x direction.) (a) what is the impulse delivered to the ball by the tennis racket?

1 answer:

morpeh [17]3 years ago

8 0

I need an answer choice

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An electron moving with a velocity = 5.0 × 10 7 m/s enters a region of space where perpendicular electric and a magnetic fields

inna [77]

Answer:

The magnetic field is $2 \times 10^{-4}$ T

Explanation:

Given:

Velocity of electron $v = 5 \times 10^{7} \frac{m}{s}$

Electric field $E = 10^{4} \frac{V}{m}$

The force on electron in magnetic field is given by,

$F = qvB \sin \theta$ ......(1)

The force on electron in electric field is given by,

$F = qE$ ......(2)

Compare both equation,

$qE = qvB \sin \theta$

Here $\sin \theta = 1$

$E = vB$

$B= \frac{E}{v}$

$B = \frac{10^{4} }{5 \times 10^{7} }$

$B = 2 \times 10^{-4}$ T

Therefore, the magnetic field is $2 \times 10^{-4}$ T

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3 years ago

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Answer:

maybe ill be tracer

Explanation:

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3 years ago

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

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$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

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3 years ago

Help with this question please.

sergij07 [2.7K]

Answer: Yes Because it matches with the mass and the amount of force Hope this helps :>

Explanation:

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algol13

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2 years ago

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