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madreJ [45]
2 years ago
6

Two foam peanuts become equally charged as they fall out of a cardboard box. Each foam peanut carries a charge of 8 x 10^-4 C. I

f the peanuts are 0.05 m apart, what is the electric force between them? Assume that k = 9 x 10^9 Nm^2/C^2. Use Coulomb’s law equation: F = kq1q2/r^2.
A. 115,200 N
B. 2,304,000 N
C. 0.000000434 N
D. 0.0000086 N
Physics
1 answer:
jok3333 [9.3K]2 years ago
7 0

Answer:

Answer labeled B : 2,304,000 N

Explanation:

Recall the formula for the electric force:

F_e=9\,\,10^9\,  \,  \frac{q_1*q_2}{d^2} \\F_e=9\,\,10^9\,  \,  \frac{8\,10^{-4}*8\,10^{-4}}{0.05^2} \\F_e=2,304,000\,\,N

which agrees with answer B provided.

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b. slow-moving streams.

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2 years ago
When a 20-V emf is placed across two resistors in series, a current of 2.0 A is present in each of the resistors. When the same
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Answer:

7.24 ohm

Explanation:

Let R1 and R2 are resistance of two resistors.

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We have to find the magnitude of the greater of the two resistances.

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R=R_1+R_2

V=IR

By using the formula

20=2(R_1+R_2)

R_1+R_2=\frac{20}{2}=10...(1)

In parallel

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}

\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}

R=\frac{R_1R_2}{R_1+R_2}

20=10(\frac{R_1R_2}{R_1+R_2}

2=\frac{R_1R_2}{10}

R_1R_2=20

R_2=\frac{20}{R_1}

Substitute the value

\frac{20}{R_1}+R_1=10

R^2_1+20=10R_1

R^2_1-10R_1+20=0

R_1=\frac{10\pm\sqrt{(-10)^2-4(20)}}{2}

By using quadratic formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

R_1=\frac{10\pm 2\sqrt 5}{2}

R_1=\frac{10+2\sqrt 5}{2}=5+\sqrt 5=7.24 ohm

R_1=\frac{10-2\sqrt 5}{2}=2.76 ohm

Substitute the value

R_2=\frac{20}{7.24}=2.76 ohm

R_2=\frac{20}{2.76}=7.24 ohm

Hence, the magnitude of the greater of the two resistance=7.24 ohm

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Answer:

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