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Finger [1]
3 years ago
10

Choose the COST tab and check that 1 day is selected. Set the Cost of electricity to 10.0 ¢/kWh. The daily cost is equal to the

daily consumption (in kilowatt-hours) multiplied by the cost per kilowatt hour (¢/kWh). What is the daily cost of an incandescent lamp?
Physics
1 answer:
AveGali [126]3 years ago
6 0

Answer:

Daily Cost = 24 ¢

Explanation:

Consider a typical incandescent bulb having a power rating of 100 W. Now, we calculate the daily consumption by multiplying the Power with time:

Daily Consumption = (Power Rating)(Time in hours)

Daily Consumption = (0.1 KW)(24 h/1 day)

Daily Consumption = 2.4 KWh

Now, we will calculate the daily cost of the incandescent lamp by multiplying the daily consumption with the unit cost.

Daily Cost = (Daily Consumption)(Unit Cost)

Daily Cost = (2.4 KWh)(10 ¢/KWh)

<u>Daily Cost = 24 ¢</u>

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Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

8 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Natasha_Volkova [10]

Ok srry bout last time but the answer is A) kelvin: time the reason is because everything else is a SI. Hope this helps and TURTLE

7 0
3 years ago
Read 2 more answers
Compare the catching of two different water balloons.
Stels [109]

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B:  Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = F_{average} × Δt = mΔV

∴ F_{average} = m·ΔV/Δt

∴ For Case A F_{average} = 149.55×8/Δt =  1196.4/Δt N

For Case B  F_{average} = 598.2×8/Δt =  4785.6/Δt

Where Δt is the same for Case A and Case B,  F_{average}  for Case B >>  F_{average}  for Case B

Therefore, Case B involves the greatest force.

4 0
3 years ago
Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This
vova2212 [387]

Answer:

1.554\times 10^{32}\ \text{kg}

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = 6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2

Radius of orbit is given by

R=\dfrac{vT}{2\pi}

We have the relation

\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}

The mass of each star is 1.554\times 10^{32}\ \text{kg}

6 0
3 years ago
The depth of the Pacific Ocean in the Mariana Trench is 36,198 ft. What is the gauge pressure at this depth
FinnZ [79.3K]

Answer:

the pressure at the depth is 1.08 × 10^{8} Pa

Explanation:

The pressure at the depth is given by,

P = h \rho g

Where, P = pressure at the depth

h = depth of the Pacific Ocean in the Mariana Trench = 36,198 ft = 11033.15 meter

\rho = density of water = 1000 \frac{kg}{m^{3} }

g = acceleration due to gravity ≈ 9.8 \frac{m}{s^{2} }

P = 11033.15 × 9.8 × 1000

P = 1.08 × 10^{8} Pa

Thus, the pressure at the depth is 1.08 × 10^{8} Pa

4 0
3 years ago
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