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Lena [83]
3 years ago
10

Which fields of engineering use fluid power? Explain how these fields make use of fluid power systems: water supply, agricultura

l equipment, construction, and earth-moving equipment.
Engineering
1 answer:
inysia [295]3 years ago
4 0

Answer:

Some fields of engineering that use fluid power are construction, heavy equipment, metallurgy, manufacturing, and oil and gas.

Water supply systems consist of these hydraulic components:

Pipes: Pipes transport water around the system. They are primary feeders (large-diameter pipes), secondary feeders (medium-diameter pipes), and distributors (small-diameter pipes).

Valves: They control the direction of the water flow. Indicating valves convey whether they are open or closed and non-indicating valves do not.

Hydrants: Hydrants appear at periodic intervals along the underground pipe network to provide water for firefighting.

Storage containers: Tanks located at a height provide water when demand is higher than supply. Gravity provides the required water pressure. Pumps can elevate water and refill the tank.

Agricultural equipment: Machines with hydraulic conveyors help transport harvested crops to wagon trailers.

In the construction industry, industrial forklifts lift and transport heavy weights. Hydraulic cylinders can tilt, lower, or elevate the required weight in a smooth motion.

Earth-moving equipment such as excavators have hydraulically powered buckets that can dig up ground, carry soil in the bucket, and drop the soil into trucks.

Explanation:

PLATO answer

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PLEASE HELP!! Its easy!!!
Rina8888 [55]

Answer:

C is tire

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4 0
3 years ago
Read 2 more answers
A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the mome
Vikki [24]

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

                           mag(BC) = sqrt ( 48^2 + 36^2) = 60 in

                           BE / BC = 45 / 60 = 0.75

Hence,                E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in

- Find unit vector EF:

                           mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

                           vec(EF) = < -15 , -110 , 30 >

                           unit(EF) = (1/115) * < -15 , -110 , 30 >

- Tension            T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb

- Find unit vector AD:

                           mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

                           vec(AD) = < 48 , -12 , 36 >

                           unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >

                           unit (AD) = <0.7845 , -0.19612 , 0.58835 >

Next:

                           M_AD = unit(AD) . ( E x T_EF)

                           M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right]

                            M_AD = 1835.73 + 116.49528 - 593.0568

                            M_AD = 1359.17 lb-in

3 0
3 years ago
A sensor produces a signal with amplitude 15 mV. A voltage amplifier must amplify the signal such that the amplitude of the outp
Nady [450]

Answer:

42.50 dB

Explanation:

Determine the minimum voltage gain

amplitude of input signal ( Vi ) = 15 mV

amplitude of output signal ( Vo) = 2 V

Vo = 2 v

therefore ; minimum gain = Vo / Vi =  2 / ( 15 * 10^-3 )

                                                         = 133.33

Minimum gain in DB = 20 log ( 133.33 )

                                  = 42.498 ≈ 42.50 dB

8 0
3 years ago
Subject: Electronics
Maslowich

Answer:

U just believe in yourself ..........

Explanation:

<em>If </em><em>there </em><em>a</em><em>r</em><em>e </em><em>more </em><em>electrons </em><em>than </em><em>protons </em><em>in </em><em>a </em><em>piece </em><em>of </em><em>matter </em><em>it </em><em>will </em><em>have </em><em>a </em><em>negative</em><em> </em><em>charge </em><em>.</em><em> </em><em>i</em><em>f</em><em> </em><em>there </em><em>are </em><em>fever </em><em>it </em><em>will </em><em>have </em><em>positive</em><em> </em><em>charge </em><em>and </em><em>if </em><em>there </em><em>are </em><em>e</em><em>qual </em><em>numbers </em><em>it </em><em>will </em><em>have </em><em>neutral</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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hope it was helpful to you.....

6 0
2 years ago
Which of the following explains why material properties present challenges for engineers?
Maurinko [17]

Answer:

Explanation:

They are altered by variables such as temperature hence making materials challenging when dealing with them.

4 0
4 years ago
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