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elixir [45]
3 years ago
9

How many kWh of energy does a 550-W toaster use in the morning if it is in operation for a total of 5.0 min? At a cost of 9.0 ce

nts/k Wh, estimate how much this would add to your monthly electric energy bill if you made toast four mornings per week.
Physics
1 answer:
iren [92.7K]3 years ago
3 0

Answer:

0.0458 kWh

6.5736 cents

Explanation:

The formula for electric energy is given as

E = Pt................. Equation 1

Where E = Electric energy, P = Electric power, t = time.

Given; P = 550 W, t = 5 min = (5/60) h = 0.083 h.

Substituting into equation 1

E = 550(0.083)

E = 45.83 Wh

E = (45.83/1000) kWh

E = 0.0458 kWh.

Hence the kWh = 0.0458 kWh.

If the makes a toast four morning per week, and the are Four weeks in a month.

Total number days he makes toast in a month = 4×4 = 16 days.

t = 16×0.083 h = 1.328 h.

Total energy used in a month = 550(1.328)

E = 730.4 Wh

E = 0.7304 kWh.

If the cost of energy is 9.0 cents per kWh,

Then for 0.7304 kWh  the cost will be 9.0(0.7304) = 6.5736 cents.

Hence this would add 6.5736 cents to his monthly electric bill

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A solenoid coil with 22 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 25.0 cm long an
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Answer:

a) 1.34*10^-8 W

b) 1.18*10^-5 H

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Explanation:

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mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A

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I: current of the inner solenoid = 0.100A

A: area of the inner solenoid = pi*r^2

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You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:

A=\pi(10^{-2}m)^2=3.141510^{-4}m^2\\\Phi_B=(4\pi*10^{-7}T/A)(340)(0.100A)(3.1415*10^{-4}m^2)=1.34*10^{-8}W\\

the magnetic flux is 1.34*10^{-8}W

b) the mutual inductance is given by:

M=\mu_o N_1 N_2 \frac{A_2}{l}

N1: turns of the outer solenoid = 22

N2: turns of the inner solenoid

A_2: area of the inner solenoid

l: length of the solenoids = 25.0cm=0.25m

by replacing all these values you obtain:

M=(4\pi*10^{-7}T/A)(340)(22)\frac{3.14*10^{-4}m^2}{0.25m}=1.18*10^{-5}H

the mutual inductance is 1.18*10^{-5}H

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\epsilon_1=M\frac{dI_2}{dt}

by replacing you obtain:

\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV

the emf is 20mV

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