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Feliz [49]
3 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]3 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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Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo
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Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567  x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

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Explanation:

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Answer:

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Answer:

Explanation:

Given that, the pilot can withstand 9g acceleration which is approximately 88m/s².

Now, the pilot is traveling in a circle of radius

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Then, acceleration?

The acceleration of a circular motion can be determine using centripetal acceleration

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