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3 years ago

7

A circuit has a voltage drop of 120 V across an 80 resistor that carries a current of 0.75 A. How much power is used by the resi

stor? Use P = VI.
A. 90 W
B. 60 W
C. 9600 W
D. 2 W

1 answer:

Pani-rosa [81]3 years ago

5 0

Answer:

A. 90 W

Explanation:

Just took the quiz

May 28, 2019

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Two parallel square metal plates that are 1.5 cm and 22 cm on each side carry equal but opposite charges uniformly spread out ov

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Answer:

The number of excess electrons are on the negative surface is $4.80\times10^{10}\ electrons$

Explanation:

Given that,

Distance =1.5 cm

Side = 22 cm

Electric field = 18000 N/C

We need to calculate the capacitance in the metal plates

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times(22\times10^{-2})^2}{1.5\times10^{-2}}$

$C=0.285\times10^{-10}\ F$

We need to calculate the potential

Using formula of potential

$V=Ed$

Put the value into the formula

$V=18000\times1.5\times10^{-2}\ V$

$V=270\ V$

We need to calculate the charge

Using formula of charge

$Q=CV$

Put the value into the formula

$Q=0.285\times10^{-10}\times270$

$Q=76.95\times10^{-10}\ C$

Here, the charge on both the positive and negative plates

$Q=+76.95\times10^{-10}\ C$

$Q=-76.95\times10^{-10}\ C$

We need to calculate the number of excess electrons are on the negative surface

Using formula of number of electrons

$n=\dfrac{q}{e}$

Put the value into the formula

$n=\dfrac{76.95\times10^{-10}}{1.6\times10^{-19}}$

$n=4.80\times10^{10}\ electrons$

Hence, The number of excess electrons are on the negative surface is $4.80\times10^{10}\ electrons$

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Is this statement true or false? Electric currents flow easily through materials that are conductors and through materials that

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Black holes are highly condensed remnants of stars. Some black holes, together with a normal star, form binary systems. In such

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$\lambda_\text{max} = 2.63\times 10^{-9}\;\text{m}$ .

<h3>Explanation</h3>

The peak emission wavelength of an object depends on its absolute temperature.

$\lambda_\text{max} = \dfrac{2.90\times 10^{-3}}{T}$ ,

where

$\lambda_\text{max}$ is the wavelength in meters where the emission is the strongest.
$T$ is the temperature of the object in degrees Kelvins.

For the gas falling into the black hole,

$T = 1.10\times 10^{6}\;\text{K}$ .

Apply the formula:

$\lambda_\text{max} = \dfrac{2.90\times 10^{-3}}{T} = \dfrac{2.90\times 10^{-3}}{1.10\times 10^{6}} = 2.64\times 10^{-9}\;\text{m} = 2.64 \;\text{nm}$ .

The question mentioned that $\lambda_\text{max}$ is in the X-ray region of the electromagnetic spectrum. According to Encyclopedia Britannica, the wavelength of X-rays range from $10^{-8}\;\text{m}=10\;\text{nm}$ to $10^{-10}\;\text{m} = 0.1\;\text{nm}$ , which indeed includes $2.64\times 10^{-9}\;\text{m} = 2.64 \;\text{nm}$ .

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