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Rzqust [24]
3 years ago
7

A single household circuit is connected to three electrical outlets (all in parallel). The outlets are connected to three device

s: an 800-W electric kettle, a 60-W trouble light, and a 1,000-W toaster oven. If all three appliances operate at the same time, will the 15-amp fuse of the circuit trip? yes or no ?
Physics
2 answers:
natima [27]3 years ago
8 0

<span>Power (Watts) = Voltage (Volts) times the Current in (Amps). 
</span>so
<span>P=I*E Or
 (P/E) = I Or (P/I) = E </span>
<span>So a 2,000 W dryer operating at 240V will draw 8.3 Amps. </span>
<span>But a 2,000 W dryer operating at 120V will draw 16.7 Amps. </span>
so
<span> 800W + 60W +1,000W total 1,860W 
</span>applying formula
<span>Using (P/E) = I </span>
<span>1,860W/120V = 15.5A </span>
<span>yep, it could trip a 15.0 A breaker
hope it helps</span>
marishachu [46]3 years ago
7 0
The total power consumption when all three of them are turned on is 1,860 watts. Power=(voltage)x(current). If the service to the household is 120 volts, then the current will be 1860/120=15.5A, which the fuse or breaker will not allow, and they will blow/trip. If the house is in South America or Europe, and the mains service is 240 volts, then the current will be 1860/240= 7.75 Amp, and the fuse or breaker will be 'cool' with that.
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Answer:

1. T₁ is approximately 100.33 N

T₂ is approximately -51.674 N

2. 230°F is 383.15 K

3. Part A

The total torque on the bolt is -4.2 N·m

Part B

Negative anticlockwise

Explanation:

1. The given horizontal force = 86 N

The direction of the given 86 N force = To the left (negative) and along the x-axis

(The magnitude and direction of the 86 N force = -86·i)

The state of the system of forces = In equilibrium

The angle of elevation of the direction of the force T₁ = 31° above the x-axis

The direction of the force T₂ = Downwards, along the y-axis (Perpendicular to the x-axis)

Given that the system is in equilibrium, we have;

At equilibrium, the sum of the horizontal forces = 0

Therefore;

T₁ × cos(31°) - 86 = 0

T₁ = 86/(cos(31°)) ≈ 100.33

T₁ ≈ 100.33 N

Similarly, at equilibrium, the sum of the vertical forces = 0

∴ T₁×sin(31°) + T₂ = 0

Which gives;

100.33 × sin(31°) + T₂ = 0

T₂ = -100.33 × sin(31°) ≈ -51.674

T₂ ≈-51.674 N

2. 230° F to Kelvin

To convert degrees Fahrenheit (°F) to K, we use;

Degrees \ in  \ Kelvin, K = (x^{\circ} F + 459.67) \times \dfrac{5}{9}

Pluggining in the given temperature value gives;

Degrees \ in  \ Kelvin, K = (230^{\circ} F + 459.67) \times \dfrac{5}{9} = 383.15

230°F = 383.15 K

3. Part A

Torque = Force × perpendicular distance from the line of action of the force

Therefore, the clockwise torque = 9 N × 0.4 m = 3.6 N·m (clocwise)

The anticlockeisre torque = 13 N × 0.6 m = 7.8 N·m (anticlockwise)

The total torque o the bolt = 3.6 N·m - 7.8 N·m = -4.2 N·m (clockwise) = 4.2 N·m anticlockwise

Part B

The torque is negative anticlockwise.

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