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3 years ago

9

Choose all statements that are true about a GFCI outlet.

1 answer:

saveliy_v [14]3 years ago

5 0

GFCI outlets are found in wet areas is the true statement about a GFCI oulet

Answer: Option D.

<u>Explanation: </u>

GFCI stands for Ground Fault Circuit Interrupter and also named as RCD (Residual Current Device). This is a kind of circuit breaker that cuts off electricity after detecting an imbalance between output and input currents. The switch protects household lines and sockets against overheating and possible fire.

GFCI protects the people and is mostly used in bathrooms or kitchen where electrical equipment is used. Also, it is available where the human body can get into contact with the floor or metal fixings that provide an alternative route to power in the event of a fault. These outlets are available in two versions: the circuit breaker installed in the control panel and receptacle type installed into the electrical box.

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The new velocity after 4 s is 40 m/s

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The given parameters for the first question;

initial velocity of the car, u = 76 m/s

acceleration of the car, a = - 9 m/s²

time of motion, t = 4 s

The new velocity after 4 s is calculated as;

v = u + at

v = 76 + (-9)(4)

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(5)

The given parameters;

height above the ground, h = 500 m

velocity of spaceship, u = 150 m/s

time of motion, t = 5

The height of the spaceship above the ground after 5 seconds is calculated as;

$h_y = h_0 + ut - \frac{1}{2}gt^2\\\\h_y = 500 + (150\times 5) - (0.5\times 9.8 \times 5^2)\\\\h_y = 1,127.5 \ m$

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2 years ago

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3 years ago

1. A huge spool of wire, 10,000 meters long, weighs 81.34 N. You cut off a meter or so and tie it between two posts, 0.660 m apa

WARRIOR [948]

Answer:

The beat frequency is 6.378 Hz.

Explanation:

Given that,

Length of wire = 10000 m

Weight = 81.34 N

Distance = 0.660 m

Tension = 52 N

Frequency = 196 Hz

We need to calculate the mass of the wire

Using formula of weight

$W= mg$

$m = \dfrac{W}{g}$

Put the value into the formula

$m=\dfrac{81.34}{9.8}$

$m=8.3\ kg$

The mass per unit length of the wire

$m_{l}=\dfrac{m}{l}$

Put the value into the formula

$m_{l}=\dfrac{8.3}{10000}$

$m_{l}=8.3\times10^{-4}\ kg/m$

We need to calculate the frequency in the wire

Using formula of frequency

$f_{w}=\dfrac{1}{2l}\sqrt{\dfrac{T}{m_{l}}}$

Put the value into the formula

$f_{w}=\dfrac{1}{2\times0.660 }\sqrt{\dfrac{52}{8.3\times10^{-4}}}$

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Using formula of beat frequency

$f_{b}=f_{in}-f_{w}$

Put the value into the formula

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