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olga2289 [7]
3 years ago
14

You are an electrician installing the wiring in a new home. The homeowner desires that a ceiling fan with light kits be installe

d in five different rooms. Each fan contains a light kit that can accommodate four 60-watt lamps. Each fan motor draws a current of 1.8 amperes when operated on high speed. It is assumed that each fan can operate more than three hours at a time and therefore must be considered a continuous-duty device. The fans are to be connected to a 15-ampere circuit. Because the devices are continuous-duty, the circuit current must be limited to 80% of the continuous connected load. How many fans can be connected to a single 15-ampere circuit
Physics
1 answer:
In-s [12.5K]3 years ago
8 0

Answer:

3 fans per 15 A circuit

Explanation:

From the question and the data given, the light load let fan would have been

(60 * 4)/120 = 240/120 = 2 A.

Next, we add the current of the fan motor to it, so,

2 A + 1.8 A = 3.8 A.

Since the devices are continuos duty and the circuit current must be limited to 80%, then the Breaker load max would be

0.8 * 15 A = 12 A.

Now, we can get the number if fans, which will be

12 A/ 3.8 A = 3.16 fans, or approximately, 3 fans per 15 A circuit.

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Radiator. A radiator puts warm air out at the top and draws in cooler air at the bottom. steaming cup of hot tea - The steam you see when drinking a cup of hot tea indicates that heat is being transferred into the air. ice melting/Ice melts because heat moves to the ice from the air
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Couldn’t you technically make infinite speed by putting a car in a vacuum chamber? Since top speed it made by the amount of forc
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No

Explanation:

For infinite speed to be achevied, one must have no sink of energy to spend. The source of entropy in this example, is the tires hitting the surface, producing heat and friction. Not to mention that you'd still need fuel to start the car, and an infinite tunnel or track, which would be impossible and speed up to process of energy loss through entropy quicker.

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3 years ago
A car speeds up from 12.0 m/s to 16.0 m/s in 8.00s what is the acceleration
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Answer:

0.5m/s²

Explanation:

acceleration =v-u/t

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=4/8

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2 years ago
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What is the common property of the elements pinctogen
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5 0
2 years ago
Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A
boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

(c) The average force on the right sled is 922.625 N

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴  22.7 × v₁'  = -3.63 × 3.05

v₁' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law,  m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴  22.7 × v₂'  = -3.63 × 3.05

v₂' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = F_{average} × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}

∴ F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s}  = 922.625 \ kg\cdot m/s^2 = 922.625 \ N  

The average force on the right sled applied by the cat while landing, \mathbf{F_{average}} = 922.625 N

Learn more about conservation of momentum here:

brainly.com/question/7538238

brainly.com/question/20568685

brainly.com/question/22257327

8 0
2 years ago
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