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3 years ago

Select three mechanical processes used in the manufacturing phase of engineering.

2 answers:

8 0

Size reduction if I’m wrong pls tell me don’t yell at me like “ITS WRONG I GOT A 20” like bro that’s ur fault so pls don’t do that

77julia77 [94]3 years ago

7 0

Answer:Size reduction

Cooling

Mixing

Explanation:

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In successive an object moves from start position, then moves 1ft, 4ft and 8ft. This is an example of a. non-uniform motion b. u

Nikolay [14]

Answer:Non-uniform motion

Explanation:

This is an example of Non-uniform motion because unequal distance traveled by an object in equal interval of time is termed as Non-Uniform motion and here also the distance traveled by object is 1 ft ,4 ft and 8 ft which is different from each other.

In Uniform motion distance traveled is equal in equal interval of time .

6 0

3 years ago

A section of highway has a free-flow speed of 55 mph and a capacity of 3300 veh/hr. In a given hour, 2100 vehicles were counted

alexira [117]

Answer:

Space mean speed = 44 mi/h

Explanation:

Using Greenshield's linear model

q = Uf ( D - $D^{2}$ /Dj )

qcap = capacity flow that gives Dcap

Dcap = Dj/2

qcap = Uf. Dj/4

Where

U = space mean speed

Uf = free flow speed

D = density

Dj = jam density

now,

Dj = 4 × 3300/55

= 240v/h

q = Dj ( U - $U^{2}$ /Uf)

2100 = 240 ( U - $U^{2}$ /55)

Solve for U

U = 44m/h

5 0

3 years ago

Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the

ra1l [238]

The evaporation rate of the n-Hexane is $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

<u>Explanation</u>:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

Where the molar flux is provided by,

$N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}$

$\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}$

$P_{t}=1 a t m=101325 P a\\$

$\text { so, } P_{A 1}=$ the vapor pressure at hexane $25 \mathrm{C}$ $=20158.2 \mathrm{Pa}$

For wind, assume negligible hexane is present, hence $P_{A 2}=0$

Now,

$\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}$

$\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}$

$P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\$

$=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}$

$=90873.57 \mathrm{Pa}$

$R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}$

$z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\$

where T = 298 K

substituting all in the equation, we get

$\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}$

$=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\$

Now,Flux $\times$ area = Molar rate of evaporation

Evaporation rate = $0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)$

Evaporation rate = $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

6 0

3 years ago

Discuss typical advantages and disadvantages of an irrigation system?

trapecia [35]

Advantages include low costs and minimal labor.Water stays in the root zone, and foliage stays dry. Drawbacks to surface irrigation include potential overwatering and wasteful runoff.

4 0

3 years ago

A ball bearing has been selected with the bore size specified in the catolog as 35.000 mm to 35.020 mm. Specify appropriate mini

Fofino [41]

Answer:

the minimum shaft diameter is 35.026 mm

the maximum shaft diameter is 35.042mm

Explanation:

Given data;

D-maximum = 35.020mm and d-minimum = 35.000mm

we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6

so From table, Selection of International Trade Grades metric series

the grade tolerance are;

ΔD = IT7(0.025 mm)

Δd = IT6(0.016 mm)

Also from Table "Fundamental Deviations for Shafts" metric series

Sf = 0.026

D-maximum

Dmax = d + Sf + Δd

we substitute

Dmax = 35 + 0.026 + 0.016

Dmax = 35.042 mm

therefore the maximum diameter of shaft is 35.042mm

d-minimum

Dmin = d + Sf

Dmin = 35 + 0.026

Dmin = 35.026 mm

therefore the minimum diameter of shaft is 35.026 mm

8 0

3 years ago