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2 years ago

Choose the three questions that an engineer should ask himself or herself when identifying the need of a problem.

Engineering

2 answers:

boyakko [2]2 years ago

5 0

Answer:

What is the goal? What will a successful solution look like? What is it that the client expects me to accomplish?

Explanation:

alexira [117]2 years ago

3 0

Answer:

What is the goal? What will a successful solution look like? What is it that the client expects me to accomplish?

Explanation:

You might be interested in

Which option shows the most valuable metallic properties

Rina8888 [55]

Malleable and ductile

non metals like plastic also have other properties but can't be malleable and ductile so they r most valuable metallic properties

6 0

1 year ago

A fill covering a wide area is to be placed at the surface of this profile. The fill has a total unit weight of 20 kN/m^3 and is

goblinko [34]

Answer:

hello your question lacks some information attached below is the complete question with the required information

answer : 81.63 mm

Explanation:

settlement of the surface due to compression of the clay ( new consolidated )

= 81.63 mm

attached below is a detailed solution to the given problem

8 0

3 years ago

6) A deep underground cavern Contains 980 cuft

Elza [17]

Answer:

15625 moles of methane is present in this gas deposit

Explanation:

As we know,

PV = nRT

P = Pressure = 230 psia = 1585.79 kPA

V = Volume = 980 cuft = 27750.5 Liters

n = number of moles

R = ideal gas constant = 8.315

T = Temperature = 150°F = 338.706 Kelvin

Substituting the given values, we get -

1585.79 kPA * 27750.5 Liters = n * 8.315 * 338.706 Kelvin

n = (1585.79*27750.5)/(8.315 * 338.706) = 15625

3 0

2 years ago

If anyone knows manufacturing plz help

sveta [45]

Answer:

I don't know ask my dad he would

Explanation:

but I can't ask him because he went to get milk and forgot to come back

8 0

3 years ago

Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o

jekas [21]

Complete Question

Complete Question is attached below.

Answer:

$V'=5m/s$

Explanation:

From the question we are told that:

Diameter $d=0.10m$

Power $P=4.0kW$

Head loss $\mu=10m$

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu$

$\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10$

$H_m=(\frac{200*10^3}{1000*9.8}-10)$

$H_m=10.39m$

Generally the equation for Power is mathematically given by

$P=\rho gQH_m$

Therefore

$Q=\frac{P}{\rho g H_m}$

$Q=\frac{4*10^4}{1000*9.81*10.9}$

$Q=0.03935m^3/sec$

Since

$Q=AV'$

Where

$A=\pi r^2\\A=3.142 (0.05)^2$

$A=7.85*10^{-3}$

Therefore

$V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}$

$V'=5m/s$

5 0

3 years ago