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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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An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass
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Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia = \frac{\pi}{64 (D_O^2 -D_i^2)}

I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4

Area = \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2

radius = \sqrt{\frac{I}{A}}

r = \sqrt{\frac{3.894}{3.015}

r = 1.136 in

slenderness ratio = \frac{L}{r}

                              = \frac{14 *12}{1.136} = 147.8

buckling load = P_cr = \frac{\pi^2 EI}}{l^2}

P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}

P_{cr} = 13.62 kips

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Air at 25 C and 1 atm is flowing over a long flat plate with a velocity of 8 m/s. Determine the distance from the leading edge o
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Answer:

Explanation:

Given data

Temprature of air=25^{\circ}

pressure =1 atm

velocity(V)=8m/s

From the table for air at 25^{\circ} and 1 atm

kinematic viscosity\left ( \nu\right )=1.562\times 10^{-5} m^{2}/s

Reynolds number for turbulent flow=5\times 10^{5}

and Re.no.=\frac{V \times X}{\nu }

therefore length where turbulent flows start is

X=\frac{\left ( Re.no.\right )\times \nu }{V}

X=0.976 m

Thickness of boundary layer is given by

\delta _x=\frac{5X}{\sqrt{Re_x}}

\delta _x=\frac{5\times 0.976}{\sqrt{5\times 10^5}}

\delta _x=6.901mm

for water

kinematic viscosity\left ( \nu\right )=8.91\times 10^{-7} m^{2}/s

Reynolds number for turbulent flow=5\times 10^{5}

and Re.no.=\frac{V \times X}{\nu }

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X=\frac{\left ( Re.no.\right )\times \nu }{V}

X=0.055m

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\delta _x=\frac{5\times 0.055}{\sqrt{5\times 10^5}}

\delta _x=0.3889mm

5 0
3 years ago
Read 2 more answers
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