Which of thWhich of the following is a true statement?

Physics

2 answers:

Mars2501 [29]3 years ago

8 0

Answer:

B. Both electric fields and forces decrease in strength as the distance increases.

Explanation:

from the Coulomb's law

$F=\frac{k q_{1} q_{2} }{r^{2}}$

electrif field is also expressed as

F = qE., so

$E = \frac{k q}{r^{2}}$

it can seen that electric force and electric field and dependent on distance. that is inverse square of distance.

so, if the electric force and eletric field decreases in strength, it is because the distance of separation has increased.

IgorC [24]3 years ago

7 0

Answer:

B. Both electric fields and forces ...

Explanation:

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Three resistors are wired in parallel with a battery. Two of the resistors have resistances of 38.7 Q/ and 89.5 Q. The current i

Lina20 [59]

Answer:

$214.9 \Omega$

Explanation:

The three resistors are connected in parallel: this means that the potential difference across each resistor is the same as the voltage of the battery. This can be calculated using the information about the $38.7 \Omega$ resistor: in fact, since we know its resistance and the current flowing through it (0.155 A), we can find the potential difference across this resistor, which is equal to the voltage of the battery:

$V=IR=(0.155 A)(38.7 \Omega)=6.0 V$

We also know the total current in the circuit, 0.250 A. This means that we can find the total resistance of the circuit, using Ohm's law:

$R_{eq}=\frac{V}{I}=\frac{6.0 V}{0.250 A}=24 \Omega$

So now we now the total resistance and the resistance of two of the 3 resistors; therefore, we can find the resistance of the 3rd resistor:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\frac{1}{R_3}=\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{24 \Omega}-\frac{1}{38.7\Omega}-\frac{1}{89.5\Omega}=0.00465 \Omega^{-1}\\R_3=\frac{1}{0.00465 \Omega^{-1}}=214.9 \Omega$