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Marina CMI [18]
3 years ago
15

The insulator is the connection between the grounded circuit conductor and the equipment grounding conductor at the service.

Engineering
1 answer:
Arada [10]3 years ago
4 0
The answer to your question is correct/ true
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In successive an object moves from start position, then moves 1ft, 4ft and 8ft. This is an example of a. non-uniform motion b. u
Nikolay [14]

Answer:Non-uniform motion

Explanation:

This is an example of Non-uniform motion because unequal distance traveled by an object in equal interval of time is termed as Non-Uniform motion and here also the distance traveled by object is 1 ft ,4 ft and 8 ft which is different from each other.

In Uniform motion distance traveled is equal in equal interval of time .

6 0
3 years ago
What is the power system?
Inessa [10]

Answer: An electric power system is a network of electrical components deployed to supply, transfer, and use electric power.

Explanation:

4 0
4 years ago
Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 20oC and exits at 3.0
icang [17]

Answer:

Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 20 o C and exits at 5.0 bar. The refrigerant undergoes a throttling process. Determine the temperature, in o C, and the quality of the refrigerant at the exit of the expansion valve.

Explanation:

3 0
3 years ago
How many volts of electricity would it take to power up an entire city? Take Tokyo for example. Please explain!
jek_recluse [69]

Answer:

maybe a couple million volts

Explanation:

because there are outlets you need to charge your phone maybe you car if you have a tesla and skyscrapers

3 0
3 years ago
The system is initially moving with the cable taut, the 15-kg block moving down the rough incline with a speed of 0.080 m/s, and
garik1379 [7]

Solution :

The spring is expanded by 2 times of the block when it moves down an inclined by x times.

Here, $x_1$ = 39 mm

        x_2 = 225 mm

a). From the work energy principal,

   Work forces = kinetic energy

$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$

$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$

$9.75= 7.5(V^2_2-0.08^2)$

$1.3= V^2_2-0.08^2$

$V_2=1.14\ m/s$

b). calculating the distance travelled by the block before it comes to rest.

Substitute the value of V_2 in (1),

$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$

$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$

$98.43x - 100(4x^2+0.156x)+0.048=0$

$98.43x - 400x^2-15.6x+0.048=0$

$82.83x - 400x^2+0.048=0$

$  400x^2- 82.83x-0.048=0$

x = 0.20 m

4 0
3 years ago
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