Ask question

Ask question

All categories

Misha Larkins [42]

3 years ago

6

Hen a gfci receptacle device is installed on a 20-ampere branch circuit (12 awg copper), what is the minimum volume allowance (i

n cubic inches) required for conductor fill for each conductor in the outlet box?

1 answer:

Bezzdna [24]3 years ago

4 0

Answer:

2.25in³

Explanation:

For a 12 awg conductor the minimum volume allowance as stated by the NEC is 2.25in³

See attached Table 314.16(B) from NEC 2011

You might be interested in

if the intensity of a person's voice is 4.6 x 10^-7 w/m^2 at a distance of 2.0 m, how much sound power does that person generate

777dan777 [17]

The sound power the person generated is $2.313 \times 10^{-5} \ W$ .

<h3>Area of the sound wave</h3>

The area of the sound wave is calculated as follows;

$A = 4\pi r^2\\\\A = 4 \pi \times (2)^2\\\\A = 50.272 \ m^2$

<h3>Power generated</h3>

The sound power the person generated is calculated as follows;

$P = I A\\\\P = 4.6\times 10^{-7} \ W/m^2 \ \ \times \ \ 50.272 \ m^2\\\\P = 2.313 \times 10^{-5} \ W$

Learn more about intensity of sound here: brainly.com/question/4431819

4 0

2 years ago

A motorcyclist drove 7 km at 57km/h and then another 7 km at 81 km/h. What was the average speed?

alex41 [277]

<u>Answer:</u>

<em>The average speed of the car is 66.9 km/h</em>

<u>Explanation:</u>

Here distance covered with the speed <em>57 km/h=7 km </em>

distance covered with the speed of <em>81 km/h=7 km</em>

<em>Average speed is equal to the ratio of total distance to the total time. </em>

<em>total distance= 7 + 7= 14 km </em>

<em> $time= \frac{distance}{speed}$ </em>

<em>time taken to cover the first 7 km= 7/57 h </em>

<em>time taken to cover the second part of the journey = 7/81 h </em>

<em>average speed = $14/(7/57+7/81)=(14 \times 57 \times 81)/945=66.9 km/h$ </em>

<u><em>Shortcut: </em></u>

<em>When equal distances are covered with different speeds average speed=2 ab/(a+b) where a and b are the variable speeds in the phases. </em>

7 0

3 years ago

Cooling causes a material to

e-lub [12.9K]

Answer:

whats the question?

Explanation:

4 0

3 years ago

Read 2 more answers

If a 80kg diver jumps off of a 5 m high dive into a regulation diving pool, how much should the temperature of the pool go up?

Agata [3.3K]

Answer:

The answer cannot be determined.

Explanation:

The energy of the diver when he hits the pool will be equal to its potential energy $mgh$ , and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.

The change in temperature ${\Delta}T$ then will be

${\Delta}T=\frac{{\Delta}Q}{mc} .$

Where m is the mass of water in the pool, c is the specific heat capacity of water, and ${\Delta}Q$ is the added heat which in this case is the energy of the diver.

Since we do not know the mass of the water in the pool, we cannot make this calculation.

7 0

3 years ago

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 87.4\,\frac{rad}{s}$ , $\omega = 0\,\frac{rad}{s}$ and $\theta-\theta_{o} = 435\,rad$ , the angular deceleration is:

$\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}$

$\alpha = -8.780\,\frac{rad}{s^{2}}$

Now, the time interval of the Deceleration Phase is obtained from this formula:

$\omega = \omega_{o} + \alpha \cdot t$

$t = \frac{\omega - \omega_{o}}{\alpha}$

If $\omega_{o} = 87.4\,\frac{rad}{s}$ , $\omega = 0\,\frac{rad}{s}$ and $\alpha = -8.780\,\frac{rad}{s^{2}}$ , the time interval is:

$t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }$

$t = 9.954\,s$

The total time needed for the grinding wheel before stopping is:

$t_{T} = 2.40\,s + 9.954\,s$

$t_{T} = 12.354\,s$

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

4 0

3 years ago