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Ronch [10]
2 years ago
10

Although most no. 10 wire used is solid, when would it be beneficial to use the stranded type? a. when connectors will be crimpe

d on b. screw terminals c. direct runs with few bends d. longer runs
Physics
1 answer:
enot [183]2 years ago
3 0

Although most no. 10 wire used is solid, it would be beneficial to use the stranded type when connectors will be crimped on. The correct option is a.

<h3>What is stranded wire?</h3>

Stranded wire is used for the crimped connections.

Although most no. 10 wire used is solid.  The stranded type wire must be used when the connectors will be crimped on.

The screw terminal uses multicore wire whereas Direct run wire uses solid wire. The longer run uses optical wire.

Thus, the correct option is a.

Learn more about stranded wire.

brainly.com/question/14364127

#SPJ1

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Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves f
stealth61 [152]

Answer:

20.96 h

Explanation:

The perimeter of the track is 2*pi*r = 20pi miles

In 10 hours, car B would have moved 20miles. So, when Car A leaves from point X, car B is 20pi - 20 miles from point X counter-clockwise and car A.

From here, we can express the distance of A from X like this:

xa = 3t

And the distance of B would be:

xb = 20pi - 20 - 2t

The time t where they would passed each other and put  12 miles between them would be the one where xa - xb is equal to 12:

xa - xb = 12

3t - (20pi - 20 - 2t) = 12

5t = 20 pi - 8

t = (20pi - 8)/5 = 10.96 h

Remember to add this value to the 10 hours car B had already been racing:

t = 20.96h

4 0
3 years ago
Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th
Delvig [45]

Explanation:

As the given data is as follows.

    R_{1} = 680 \ohm ohm\ohm,    R_{2} = 720 \ohm ohm,

   R_{3} = 1.2 k\ohm = 1200 \ohm   (as 1 k ohm = 1000 m)

(a)   We will calculate the maximum resistance by combining the given resistances as follows.

      Max. Resistance = R_{1} + R_{2} + R_{3}

                                  = (680 + 720 + 1200) \ohm ohm

                                  = 2600 ohm

or,                               = 2.6 k\ohm ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 k\ohm ohm.

(b)   Now, the minimum resistance is calculated as follows.

      Min. Resistance = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}

                                 = \frac{1}{680} + \frac{1}{720} + \frac{1}{1200}

                                 = 3.683 \times 10^{-3} ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is 3.683 \times 10^{-3} ohm.

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