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AfilCa [17]
2 years ago
14

A proton is located at the point (x = 1.0 nm, y = 0.0 nm) and an electron is located at the point (x = 0.0 nm, y = 4.0 nm). Find

the magnitude of the electrostatic force that each one exerts on the other. (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2, e = 1.6 × 10-19 C) A) 5.3 × 10-18 NB) 5.3 × 108 N C) 1.4 × 10-11 N D) 5.9 × 10-15 N
Physics
1 answer:
enot [183]2 years ago
7 0

Answer:

C : 1.4*10^(-11) N.

Explanation:

q_1 = q_2 = 1.6 * 10^(-19)

R^2 = (1)^2 + (4)^2 = 1.7 * 10^(-17) m^2

The coulomb's law is as follows:

F_e = k*q_1*q_2 / R^2

F_e = k*q^2 / R^2

F_e = (9.0*10^9) * (1.6 * 10^(-19))^2 / 1.7 * 10^(-17)

F_e = 1.35 * 10^(-11) N

Hence, answer closes to obtained is option C : 1.4*10^(-11) N.

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Starting from the angular velocity, we can calculate the tangential velocity of the stone:
v=\omega r= (12 rad/s)(0.5 m)= 6 m/s

Then we can calculate the angular momentum of the stone about the center of the circle, given by
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where
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v its tangential velocity
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Substituting the data of the problem, we find
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2 years ago
At what height h above the ground does the projectile have a speed of 0.5v?
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Answer:

h=\dfrac{3v^2}{8g}

Explanation:

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v=u+at

v=u-gt

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h=vt-\dfrac{1}{2}gt^2

h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2

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So, the height of the projectile above the ground is \dfrac{3v^2}{8g}. Hence, this is the required solution.

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Technician A says test lights are great for quick tests on non-computerized circuits. Technician B says you can use a test light
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that technician A is right

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molecular weights are written in the picture.

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