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WARRIOR [948]
3 years ago
12

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?
Physics
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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Travka [436]
The correct answer is
<span>c) very small and very large

Let's see this with a few examples:
1) if we have a very small number, such as
</span>0.0000000001
<span>we see that we can write it easily by using the scientific notation:
</span>1\cdot 10^{-10}
<span>2) Similarly, if we have a very large number:
</span>10000000000
<span>we see that we can write it easily by using again the scientific notation:
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4 0
3 years ago
HELP ASAP PLEASE!!!
Montano1993 [528]

Answer:

The answer is A, B, C, D

Explanation:

This is because gravity is the weakest force of the four fundamental forces, so it automatically cancels letter E

7 0
2 years ago
Read 2 more answers
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
densk [106]
  • Voltage=V=120V
  • Power=P=60W

  • Current=I

\\ \rm\rightarrowtail I=\dfrac{P}{V}=\dfrac{60}{120}=\dfrac{1}{2}=0.5A

8 0
2 years ago
Hurry up i need help.
spayn [35]

Answer:

The answer would be B) The Same

Explanation:

Not gonna lie I checked my class notes but I figured this would help :)

Good luck!!!

8 0
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.

7 0
3 years ago
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