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3 years ago

Which ancient civilization can you trace the development of manufacturing

Engineering

1 answer:

nikdorinn [45]3 years ago

8 0

Answer:

Explanation:

Sumerians invented new technologies and perfected the large-scale use of existing ones. In the process, they transformed how humans cultivated food, built dwellings, communicated and kept track of information and time

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When were dresses made

klio [65]

Answer:

The world's oldest dress called the Tarkhan Dress is at 5,100 to 5,500 years of age.

Does that help? Or do you need something else? I can change my answer if this is not what you need! :D

Explanation:

6 0

3 years ago

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic

DiKsa [7]

Answer:

The answer is below

Explanation:

1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:

$n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm$

2) The speed of the rotor is the motor speed. The slip is given by:

$Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm$

3) The frequency of the rotor is given as:

$f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz$

4) At standstill, the speed of the motor is 0, therefore the slip is 1.

The frequency of the rotor is given as:

$f_r=slip*f_s\\f_r=1*60=60\ Hz$

6 0

3 years ago

Ti-6Al-4V has a fracture toughness of 74.6 MPa-m0.5. How much stress (in MPa) would it take to fail a plate loaded in tension th

Nikitich [7]

Answer:

critical stress = 595 MPa

Explanation:

given data

fracture toughness = 74.6 MPa- $\sqrt{m}$

crack length = 10 mm

f = 1

solution

we know crack length = 10 mm

and crack length = 2a as given in figure attach

so 2a = 10

a = 5 mm

and now we get here with the help of plane strain condition , critical stress is express as

critical stress = $\frac{k}{f\sqrt{\pi a}}$ ......................1

put here value and we get

critical stress = $\frac{74.6}{1\sqrt{\pi 5\times 10^{-3}}}$

critical stress = 595 MPa

so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.

plain stress condition occur in thin body where stress through thickness not vary by the thinner section.

6 0

3 years ago

The total solids production rate in an activated sludge aeration tank is 7240 kg/d on a dry mass basis. It is necessary to maint

snow_lady [41]

Answer:

volume of biological sludge = 28.566 m³ per day

Explanation:

given data

mass of solid = 7240 kg/day

initial moisture content = 78%

solution

here percentage of solid will be

% of solid = 100 - initial moisture content

% of solid = 100 - 78 = 22 %

so that

mass of sludge produced = $\frac{100}{100 - P} M$ kg per day

put her value

mass of sludge produced = $\frac{100}{100 - 78} 7240$ kg

mass of sludge produced = 32909.09 kg

specific gravity of sludge = $\frac{\rho sludge}{\rho water }$

and as we know that

$\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}$

$\frac{100}{S sludge} = \frac{22}{2.5} = \frac{78}{1}$

$S sludge$ = 1.152

so that

density of sludge = S sludge × density of water

density of sludge = 1.152 × 1000

density of sludge = 1152 kg/m³

so that

volume of biological sludge = $\frac{mass sludge produce}{\rho sludge}$

volume of biological sludge = $\frac{32909.09}{1152}$

volume of biological sludge = 28.566 m³ per day

6 0

3 years ago

Maximum iorsional shear siress.? Select one: a)- occurs at the center of a shaft. b)- occurs at the outer surface of a shaft c)-

pickupchik [31]

Answer:

b). Occurs at the outer surface of the shaft

Explanation:

We know from shear stress and torque relationship, we know that

$\frac{T}{J}= \frac{\tau }{r}$

where, T = torque

J = polar moment of inertia of shaft

τ = torsional shear stress

r = raduis of the shaft

Therefore from the above relation we see that

$\tau = \frac{T.r}{J}$

Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.

When r= 0, then τ = 0

and when r = R , τ is maximum

Thus, torsional shear stress is maximum at the outer surface of the shaft.

4 0

3 years ago