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3 years ago

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In doing a load of clothes, a clothes dryer uses 16 A of current at 240 V for 45 min. A personal computer, in contrast, uses 2.7

A of current at 120 V. With the energy used by the clothes dryer, (a) How long (in hours) could you use this computer to surf the Internet? (b)At $0.12 per kilowatt-hour, how much does it cost to operate the clothes dryer for 45 minutes?

1 answer:

EastWind [94]3 years ago

4 0

Answer:

(a) Total time will be 8.05 hour

(B) Total cost will be $417.6

Explanation:

We have given dryer uses 16 A of current so current i = 16 A

Voltage V = 240 volt

Time t = 45 minutes =45×60 = 2700 sec

So power P = VI = 240×16 = 3840 W

So energy E = power×time = 3840×45×60 = 9396 KJ

(A) Now current in the computer i = 2.7 A

Voltage V = 120 V

So power P = 120×2.7 = 324 W

Now computer is using energy of dryer

So time by which it will use energy is $t=\frac{9396000}{324}=29000sec=8.05hour$

So time will be 8.05 hour

(b) Cost per kilowatt is $0.12

So total cost =3480×0.12 = $417.6

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2 years ago

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A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb

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2 years ago

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3 years ago

A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

Morgarella [4.7K]

Answer:

(A). The work done is $1.50\times10^{-6}\ J$ .

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy $K.E=1.50\times10^{-6}\ J$

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W= 1.50\times10^{-6}-0$

$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

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3 years ago