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Feliz [49]
2 years ago
7

A hair dryer draws 11 A when it is connected to 120 V. If electrical energy costs $ 0.09/kW·h, what is the cost of using the hai

r dryer for exactly 15 min?
Free brainliest, i was planning on answering this myself anyway.

Answer: $0.3

Explanation:
Operating Cost ($) = Energy Use (kWh) x Electric Rate
1. convert Amps to watts
11 x 120= 1320
2. convert watts to kilowatts
1320/1000=1.32
3. convert the time to hours
15/60=0.25
4. Multiply the time and the kilowatts
0.25 x 1.32= 0.33kWh
5. find the cost
0.9 x 0.33= $0.297
Physics
1 answer:
Papessa [141]2 years ago
6 0

Answer:

The cost of using the hair dryer for 15 minutes is \$3.6\bar 6

Explanation:

The parameters given in the question are;

The electric current drawn by the the air dryer, I = 11 A

The voltage to which the hair dryer is connected, V = 120 V

The duration of usage of the hair dryer = 15 minutes = 60 minutes /4 = 1 hour/4 = 0.25 hour

The electrical energy costs $0.09/kW·h

The power consumed by the hair dryer = I × V = 11 × 120 = 1320 Watts = 1.32 kW

The energy used by the hair dryer in 15 minutes (0.25 hour) = 1.32 × 0.25 0.33 kW·h

The energy used by the hair dryer in 15 minutes (0.25 hour) = 0.33 kW·h

The energy cost = $0.09/(kW·h)

Therefore;

The cost of using the hair dryer for 15 minutes (0.25 hour) = 0.33 kW·h/($0.09/(kW·h)) = $33/9 = $3 2/3 = $3.6\bar 6.

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muminat

Power P is the rate at which energy is generated or consumed and hence is measured in units that represent energy E per unit time t. This is:

P = E/t

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t = E/P

t = 6007 J / 500 W

t = 12.014 s

<h2>t ≅ 12 s</h2>

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3 years ago
Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin
Zarrin [17]

Answer:

\rm 9.186\times 10^{-7}\ C.

Explanation:

<u>Given:</u>

  • Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
  • Distance of separation between the plates, d = 1.0 cm = 0.01 m.
  • Minimum value of electric field that produces spark, \rm E=3\times 10^6\ N/C.

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

\rm E=\dfrac{\sigma}{\epsilon_o}.

where,

  • \rm \sigma = surface charge density of the plate of the capacitor = \dfrac qA.
  • \rm q = magnitude of the charge on each of the plate.
  • \rm A = surface area of each of the plate =\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.
  • \epsilon_o = electrical permittivity of free space, having value = 8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.

For the minimum value of electric field that produces spark,

\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

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