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Akimi4 [234]
3 years ago
5

Select the correct answer What is the simplest definition of a manufacturing process?

Engineering
2 answers:
anastassius [24]3 years ago
7 0

a process that converts raw materials into a useful product.

lukranit [14]3 years ago
3 0

Answer:

B. A process that mass-produces complex products :)

You might be interested in
Write a matrix, that is a lower triangular matrix.
shepuryov [24]

Answer:

\left[\begin{array}{ccc}10&0&0\\14&25&0\\57&18&39\end{array}\right]

Explanation:

A lower triangular matrix is one whose elements above the main diagonal are zero meanwhile all the main diagonals elements and below are nonzero elements. This is one of  the two existing types of triangular matrixes. Attached you will find a image referring more about triangular matrixes.

If there is any question, just let me know.

6 0
4 years ago
Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
love history [14]

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

\frac{P_{1} }{\gamma } +\frac{V_{1}^{2}  }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2}  }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}

For the pipe 1, the flow velocity is:

V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }

Q = 18 L/s = 0.018 m³/s

D = 6 cm = 0.06 m

V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4

\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

h_{1} =\frac{V_{1}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m

For the pipe 2, the flow velocity is:

V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4

\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087

The head of pipe 1 is:

h_{2} =\frac{V_{2}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m

The total head is:

hi = 1326.18 + 21.3 = 1347.48 m

The required pump head is:

h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m

The required pumping power is:

P=Q\rho *g*h_{pump}  =0.018*999.1*9.8*1344.55=236965.16W=236.96kW

8 0
3 years ago
The end of a large tubular workpart is to be faced on a NC vertical boring mill. The part has an outside diameter of 38.0 in and
nata0808 [166]

Answer:

(a) the cutting time to complete the facing operation = 11.667mins

b) the cutting speeds and metal removal rates at the beginning= 12.89in³/min and end of the cut. = 8.143in³/min

Explanation:

check attached files below for answer.

5 0
3 years ago
In the LC-3 data path, the output of the address adder goes to both the MARMUX and the PCMUX, potentially causing two very diffe
dangina [55]

Answer:

no need for that

Explanation:

they are not the same at all

3 0
3 years ago
Which term defines the amount of mechanical work an engine can do per unit of heat energy it uses?
skad [1K]

Answer:

d

Explanation:

is the because that's the amount of work in making machine can do producing heat

7 0
3 years ago
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