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Leto [7]
3 years ago
13

A particle of mass 73 g and charge 67 µC is released from rest when it is 47 cm from a second particle of charge −25 µC. Determi

ne the magnitude of the initial acceleration of the 73 g particle. Answer in units of m/s 2 .
Physics
1 answer:
stich3 [128]3 years ago
8 0

Answer:

933.804423995 m/s²

Explanation:

q_1 = Charge on particle 1 = 67 µC

q_2 = Charge on particle 2 = -25 µC

r = Distance between the particles = 47 cm

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

m = Mass of particle = 73 g

Electric force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times -25\times 10^{-6}\times 67\times 10^{-6}}{0.47^2}\\\Rightarrow F=-68.1677229516\ N

The magnitude of force is 68.1677229516 N

Acceleration is given by

a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{68.1677229516}{0.073}\\\Rightarrow a=933.804423995\ m/s^2

The acceleration is 933.804423995 m/s²

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Answer:

A, 0.59A

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To calculate the current flowing through the 10ohm resistance we need to know the voltage drop by subtracting the voltage drop in the 2ohm resistance from the total voltage drop.

Voltage drop on the 2 ohm resistance is;

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12/3.94= 3.05A

Voltage drop on the 2 ohm resistance;

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Remember V= I× R so that I = V/R

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