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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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Answer:

1) The greatest height attained by the ball equals 20.387 meters.

2) The time it takes for the ball to reach 15 meters approximately equals 1 second.

Explanation:

The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.

thus using third equation of kinematics we obtain the height attained as

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where

'v' is the final speed of the ball

'u' is the initial speed of the ball

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's' is the distance it covers

Thus for maximum height applying the values in the equation we get

0=20^{2}-2\times 9.81\times h\\\\\therefore h=\frac{20^{2}}{2\times 9.81}=20.387meters

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as  

v^2=20^{2}-2\times 9.81\times 15\\\\v^{2}=105.7\\\\\therefore v=10.28m/s

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v=u+at\\\\t=\frac{v-u}{a}\\\\t=\frac{10.28-20}{-9.81}=0.991seconds\approx 1second

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