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Gemiola [76]
2 years ago
13

Two technicians are discussing relays. Technician A says that relays can fail because the relay winding is open. Technician B sa

ys relays can fail because the switch contacts can have high resistance. Who is correct
Engineering
1 answer:
hram777 [196]2 years ago
4 0

Technician A says that relays can fail because the relay winding is open. Technician A is correct.

<h3>What is winding?</h3>

A single turn of the wound material, which is material wrapped or coiled around an object.

Poor contact alignment and open coils can also cause relays to malfunction.

The most important aspect impacting relay dependability is choosing the right relay type for a particular application. When incorporating them into circuits, several subpar design techniques are employed.

Hence, technician A is correct.

To learn more about the winding refer;

brainly.com/question/23369600

#SPJ1

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5. The water in an 8-m-diameter, 3-m-high above-ground swimming pool is to be emptied by unplugging a 3-cm-diameter, 25-m-long h
frosja888 [35]

Answer:

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Friction head and pressure head will cause the actual flow rate to be less.

Explanation:

Considering point 1 at the free surface of the pool, and point 2 at the exit of

pipe.

Using Bernoulli equation between

these two points simplifies to

P1/(p*g) + V1²/2g + z1 = P2/(p*g) + V2²/2g + z2

Let the reference level at the pipe exit (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0),

P/(p*g) + z1 = P/(p*g) + V2²/2g

z1 = V2²/2g

Note; z1 = h

V2max = √2gh

h = 3 m

V2max = √2 * 9.81 * 3

V2max = √58.86 = 7.67 m/s

maximum discharge rate of water through the pipe Qmax = Area A * Velocity of discharge V2max

Qmax = A * V2max

Diameter d = 3 cm = 0.03 m

A = Πd²/4 = (Π * 0.03²)/4 = 0.00071m³

Qmax = 0.00071 * 7.67 = 0.00545 m³/s

Qmax = 5.45 L/s

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Actual flow rate will be less because of heads such as friction head and pressure head.

7 0
3 years ago
Alternating current flows in___ direction(s) and direct current flows in___direction(s)
Oduvanchick [21]

Answer:

<u><em>both, one</em></u>

Explanation:

<em>Alternating current flows in both directions and direct current flows in one  direction.</em>

<em></em>

<em>Hope it helps.</em>

<em>;)</em>

<em><3</em>

8 0
3 years ago
A compressible clay layer has a thickness of 3.8 m. After 1.5 yr, when the clay is 50% consolidated, 7.3 cm of settlement has oc
grigory [225]

The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.

<h3>How to determine the amount of settlement?</h3>

For a layer of 3.8 m thickness, we were given the following parameters:

U = 50% = 0.5.

Sc = 7.3 cm.

For Sf, we have:

Sf = Sc/U

Sf = 7.3/0.5

Sf = 14.6

Therefore, Sf for a layer of 38 m thickness is given by:

Sf = 14.6 × 38/3.8

Sf = 146 cm.

At 50%, the time for a layer of 3.8 m thickness is: t_{50} = 1.5 year.

At 50%, the time for a layer of 38 m thickness is:

t_{50} = 1.5 × (38/3.8)²

t_{50} = 150 years.

For the thickness of 38 m, U₂ is given by:

\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25  \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05

The new settlement after 1.5 year is:

Sc = U₂Sf

Sc = 0.05 × 146

Sc = 7.3 cm.

For time, t₂ = 5 year:

U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25  \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09

The new settlement after 5 year is:

Sc = U₂Sf

Sc = 0.09 × 146

Sc = 13.14 cm.

Read more on clay layer here: brainly.com/question/22238205

8 0
2 years ago
Power cords can be damaged by Which of the following?
Elena L [17]
What are the options??
3 0
3 years ago
A rectangular channel 3.0 m wide has a flow rate of 5.0 m3/s with a normal depth of 0.50 m. The flow then encounters a dam that
Lina20 [59]

Answer:

The hydraulic will jump since the flow is subcritical ( i.e. Y2 > Yc )

Explanation:

width of channel = 3.0 m

Flow rate = 5 m^3/s

Normal depth = 0.50 m

Flow encounters a dam rise of 0.25 m

To know if the hydraulic jump will occur

we will Determine the new normal depth

Y2 = 3.77m

Yc ( critical depth )= 0.66m

Attached below is the detailed solution

8 0
3 years ago
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