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3 years ago

Which of the following would NOT be

Engineering

1 answer:

nikdorinn [45]3 years ago

7 0

Answer:

Option A

hands-on assembly of the company's product

Explanation:

The managerial roles mainly include supervision of actual work on a higher level and implementation of organization's policy. The hands-on assembly of company's product is majorly done by junior staff who work under a supervisor who reports to the manager. Therefore, this is not a duty of a manager in a manufacturing field.

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4. A 25 km2 watershed has a time of concentration of 1.6 hr. Calculate the NRCS triangular UH for a 10-minute rainfall event and

Alika [10]

Answer:

The NRCS triangular UH for a 10-minute rainfall is Qp = 49.84 m³/s

Peak flow of the aggregated runoff hydrograph is 420.58 m³/s

The total volume of runoff is 2125000 m³/s

Explanation:

We have

A = 25 km²

tr = 10 min = 1/6 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr

Qp = 2.08×25/1.043 = 49.84 m³/s

Tb = 8/3×Tp = 8/3×1.043 = 2.782 hr

Since the area is

Time (min) Runoff (cm) Volume of runoff m³

0 0 0

10 4 1000000 m³

20 2.5 625000 m³

30 2 500000 m³

Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s

For the 1st 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×4/1.043 = 197.92 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr

For the 2nd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 123.7 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr

For the 3rd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 98.96 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr

Peak flow of aggregate runoff is given by

Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s

Total volume of runoff is given by

Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s

6 0

4 years ago

Which 2 items are most similar and why? Graduated Cylinder, Measuring Cup, Beaker

aivan3 [116]

Graduated cylinder & beaker; both are used for science activities while a measuring cup is mostly used for baking.

5 0

3 years ago

A tungsten matrix with 20% porosity is infiltrated with silver. Assuming that the pores are interconnected, what is the density

daser333 [38]

Answer:

15.4 g/cm³, 17.4 g/cm³

Explanation:

The densities can be calculated using the formula below

ρ = (fraction of tungsten × ρt ( density of tungsten)) + (fraction of pores × ρp( density of pore)

fraction of tungsten = (100 - 20 ) % = 80 / 100 = 0.8

a) density of the before infiltration = ( 0.8 × 19.25) + (0.2 × 0) = 15.4 g/cm³

b) density after infiltration with silver

fraction occupied by silver = 20 / 100 = 0.2

density after infiltration with silver = ( 0.8 × 19.25) + (0.2 × 10) = 17.4 g/cm³

5 0

3 years ago

An agricultural manager requires

sp2606 [1]

C, being able to maintain legal information on grant programs

3 0

3 years ago

A steel cylindrical sample was subjected to a tensile test. The yield load was 2100N. The maximum load was 3400N and the failure

Misha Larkins [42]

Answer:

initial diameter of the sample is 2.95 mm

Explanation:

given data

yield load = 2100 N

maximum load = 3400 N

failure load = 2350 N

ultimate engineering stress = 497.4 MPa = 497 × $10^{6}$ N/m²

to find out

What was the initial diameter of the sample in mm

solution

we will apply here ultimate engineering stress formula that is express as

ultimate engineering stress = $\frac{Pmax}{A}$ ...............1

here A is area and P max is maximum load applied

so area = $\frac{\pi }{4} d^2$

here d is initial diameter

so put all value in equation 1

497 × $10^{6}$ = $\frac{3400}{\frac{\pi }{4} d^2}$

solve it we get d

d = 2.95 × $10^{-3}$ m

so initial diameter of the sample is 2.95 mm

7 0

4 years ago