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Reika [66]
3 years ago
14

1. Design a circuit, utilizing set/reset coils where PB 1 starts Motor 1 and PB2 stops Motor 1. Pressing and releasing either pu

sh button should latch the motor either on or off depending on the function

Engineering
1 answer:
lianna [129]3 years ago
5 0

Answer:

Circuit attached with explanation

Explanation:

Hi Dear,

A circuit is attached for your reference.

When you press "start" PB, the supply reaches the motor starter relay coil "M" that is also in parallel with the "start" PB which allows the motor to remain ON even when you release "start" PB as supply to relay coil is directly from supply "L" through "M".

To stop motor just press "stop" PB and the circuit breaks which de-energize the relay coil and the motor stops.

Hope this finds easy to you.

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Kitty [74]

Answer:

1. Work is a driving force of identity in your life, whether its because it influences you due to the time spent there. Or its truly a passion of yours.

2. Two main aspects of identity development are self-concept and self-esteem

3. Based on you obviously

Hope this helped!

Explanation:

5 0
3 years ago
A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Po
kicyunya [14]

This question is incomplete, the missing image in uploaded along this answer below.

Answer:

The required stress is 200 Mpa

Explanation:

Given the data in the question;

diameter D = 12 mm = 12 × 10⁻³ m

Length L = 188 mm = 188 × 10⁻³ m

Poisson's ratio v = 0.34

Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m

The transverse strain will;

εˣ = Δd / D

εˣ = -0.0105 × 10⁻³ /  12 × 10⁻³ m

εˣ = -0.00088

The longitudinal strain will be;

E^z = - ( εˣ  / v )

E^z = - ( -0.00088  / 0.34 )

E^z = - ( - 0.002588 )

E^z = 0.0026

Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.

From the graph, in the Second image;

The stress is 200 Mpa

Therefore, The required stress is 200 Mpa

8 0
3 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
You can help build a safe work environment by using your knowledge of violence prevention strategies to spot what?
Ostrovityanka [42]

Answer:

warning signs

Explanation:

give directions on your surroundings

4 0
3 years ago
What is the diffrence between a small block and a big block lets see if yall know
ivann1987 [24]

Answer:

The difference in weight and size?

Explanation:

It explains itself :P

5 0
2 years ago
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