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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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What is the rated power output in ( kw) of a 8 pole motor designed to an IEC 180L motor frame ?<br>

ra1l [238]

Answer:

P=11 kW

Explanation:

Given that

Number of poles= 8

I.E.C. 180L motor frame

From data book , for 8 poles motor at 50 Hz

Speed = 730 rpm

Power factor = 0.75

Efficiency at 100 % load= 89.3 %

Efficiency at 50 % load= 89.1 %

Output power = 11 kW

Therefore the rated output power of 8 poles motor will be 11 kW. Thus the answer will be 11 kW.

P=11 kW

4 0

3 years ago

Suppose you were doing an A* search and there were a number of different ways that you could compute an admissible heuristic val

Elodia [21]

Answer:

Explanation:

First of all,we need to know what kind of research and how important it is before determining which computation method to use .

Then compare end result of research and cost of expensive methods .

Then we should opt for the best expensive method to be used for the given application.

5 0

3 years ago

A compressible clay layer has a thickness of 3.8 m. After 1.5 yr, when the clay is 50% consolidated, 7.3 cm of settlement has oc

grigory [225]

The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.

<h3>How to determine the amount of settlement?</h3>

For a layer of 3.8 m thickness, we were given the following parameters:

U = 50% = 0.5.

Sc = 7.3 cm.

For Sf, we have:

Sf = Sc/U

Sf = 7.3/0.5

Sf = 14.6

Therefore, Sf for a layer of 38 m thickness is given by:

Sf = 14.6 × 38/3.8

Sf = 146 cm.

At 50%, the time for a layer of 3.8 m thickness is: $t_{50}$ = 1.5 year.

At 50%, the time for a layer of 38 m thickness is:

$t_{50}$ = 1.5 × (38/3.8)²

$t_{50}$ = 150 years.

For the thickness of 38 m, U₂ is given by:

$\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25 \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05$