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3 years ago

How much electrical energy is used by a 400 W toaster that is operating for 5 minutes?

Physics

1 answer:

Shkiper50 [21]3 years ago

3 0

The electrical energy consumed by a toaster is 0.033 Kwh.

Explanation:

The power utilized by the toaster is 400 W.

$\text { The power utilized by the toaster is } 400 \mathrm{W} \text { in kilo-watts is } \frac{400}{1000}=0.4 \mathrm{Kw}$

The toaster is operated for 5 Minutes.

$\text { The toaster is operated for } 5 \text { Minutes in hours is } \frac{5}{60}=0.083 \text { hours. (One minute is } 60 \text { seconds) }$

We know that,

$\text {power}=\frac{\text {energy}}{\text {time}}$

Substitute the values in the above formula to obtain electrical energy,

$0.4=\frac{\text { energy }}{0.083}$

Electrical energy = 0.4 × 0.083

Electrical energy is 0.033 Kwh.

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Answer:

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3 years ago

If themass is 50kg, what weight of water is to be displaced to float on water? why

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1 year ago

A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc

Elanso [62]

Answer:

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Explanation:

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$y_1 = \frac{1\lambda D}{d}$

$y_1 = {706*D}{d}$

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$y_m = {(m + 1/2)\lambda D}{d}$

Now to get the dark fringes at the same location we should have;

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3 years ago

Which type of exercise is weightlifting?

MatroZZZ [7]

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Nonportrit

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3 years ago

Read 2 more answers

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

3 years ago