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Vadim26 [7]
3 years ago
13

Two long pipes convey water between two reservoirs whose water surfaces are at different elevations. One pipe has a diameter twi

ce that of the other; both pipes have the same length and the same value of f. If minor losses are neglected, what is the ratio of the flow rates through the two pipes?
Engineering
1 answer:
Law Incorporation [45]3 years ago
7 0

Answer:

1 : 5.66 ≈ 1 : 6

Explanation:

Determine the ratio of the flow rates through the two pipes

The flow from the bigger pipe is 5.66 times greater than that from the smaller pipe hence the ratio of the flow rates is : 1 : 5.66

attached below is the detailed/remaining part of the  solution

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Air is compressed slowly in a piston-cylinder assembly from an initial state where P1 = 1.4 bar, V1= 4.25 m^3, to a final state
lord [1]

Answer:

W=-940.36 KJ

Explanation:

Given that

P_1=1\ bar,V_1=4.25 {m^3}

P_2=6.8\ bar

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

W=P_1V_1\ln \dfrac{P_1}{P_2}

Now by putting the values                (1.4 bar =140 KPa)

W=P_1V_1\ln \dfrac{P_1}{P_2}

W=140\times 4.25 \ln \dfrac{1.4}{6.8}

W=-940.36 KJ

Negative sign indicates that this is a compression process and work will given to the system.

7 0
4 years ago
Elijah wants to make some changes to a game that he is creating. Elijah wants to move the bleachers to the right, the coaches to
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Answer:

the editor

Explanation:

the variable is usually a number, and the conditional loop has nothign to do with movement/ where something is located

7 0
3 years ago
Read 2 more answers
What is the acceleration of a car that has a velocity of 20 m/s, and
ella [17]

Acceleration of Car = 10 ms⁻²

Explanation:

Step 1:

The basic formula of acceleration is a = (v-u)/t  ms⁻²

where, v- final velocity

            u- initial velocity

             t= time taken

Step 2:

Here v = 70 ms⁻¹

        u = 50  ms⁻¹

         t = 5 s

∴ a = ( 70 - 20)/5

a = 10 ms⁻²

7 0
4 years ago
Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows the pouring
umka2103 [35]

Answer:

the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

Explanation:

Given the data in the question ;

flowrate Q = 400 cm³/s

cross section of the sprue is round

Diameter of sprue at the top d_top = 3.4 cm

Height of sprue = 20 cm = 0.2 m³

the proper diameter at its base so as to maintain the same volume flow rate = ?

first we determine the velocity at the sprue base

V_base = √2gh = √( 2×9.81×0.2) = √3.924 = 1.980908 m = 198.0908 cm

so, diameter of the sprue at the bottom  will be

Q = AV = [ (( πd²_bottom)/4) × V_bottom ]

d_bottom =  √(4Q/πV_bottom)

we substitute

d_bottom =  √((4×400)/(π×198.0908 ))

d_bottom =  √( 1600/622.3206)

d_bottom =  √2.571022

d_bottom =  1.6034 cm

Therefore, the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

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3 years ago
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Neutral posture is essential for optimal wellbeing and functioning of the body. Holding the weight of the body The most important function of a neutral posture is to maintain the body in an upright position, supporting the body against gravity
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