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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:

Option D (N/A-m)(m)(m/s)

Explanation:

(N/A-m)(m)(m/s) is the required dimensional analysis for calculation of emi

3 0
2 years ago
To determine a waves' frequency, you must know the
givi [52]
The correct answer is<span> number of oscillations in a given period of time

This is measured in what is called the Hertz measurement and the period of time is usually taken to be per second.</span>
6 0
3 years ago
A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a
IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>
  • Range = U²× sin(2θ)/g
  • U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
  • U=√{Range×g/sin(2θ)}
  • Here, range= 2.20m, = 36.5°
  • U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>
  • Time of flight= (2×U×sinθ)/g
  • So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

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4 0
2 years ago
A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3
lesantik [10]

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

             T =  F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm = \frac{2}{100} = 0.02m for z the perpendicular distance

So Elbow Torque is   T_e= 4000 * 0.02

                                   = 80Nm

 To obtain the torque required we substitute 300 N for F and 30cm =\frac{30}{100} = 0.3 m

  So the Required Torque is T_R = 300 *0.3

                                                     =90Nm

Now since   T_e < T_R it mean that the weight lifter would not get past this sticking point

                                   

7 0
3 years ago
Potential difference is measured in which units?<br> volts<br> amps<br> currents<br> watts
viktelen [127]
It is Volt
First option
4 0
2 years ago
Read 2 more answers
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