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MrRa [10]
3 years ago
5

The development of various technologies led to many historic events. Use information from the Internet to describe one major his

toric event
initiated by a technological development or discovery

-


Sometimes, world events lead to the development of a technology. Perform online research and identify a historic event that required the

development of technologies. Discuss one technology created because of this event. How does it continue to affect human lives today?
Engineering
1 answer:
11111nata11111 [884]3 years ago
4 0

Answer:

1. Industrial revolution was initiated or borne through the production of Steel

2. World War 1 led to the development of Tanks

Explanation:

The production of Steel through the Bessemer Process in the middle of the nineteenth century was a major technological development that spurred the Industrial revolution. This invention led to the widespread use of steel in the production of many things including vehicles and airplanes.

During the First World War in 1914, soldiers found the use of just their armaments in battle as not so productive. This led to the development of Tanks in 1915 that would continue moving towards the enemy even when being shot at.

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In manufacturing a particular set of motor shafts, only shaft diameters of between 38.00 and 37.50 mm are usable. If the process
Masteriza [31]

Answer:

P(37.5

And we can find this probability with this difference:

P(-2.5

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(-2.5

So then the percentage usable are 94,6%

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the diameters of a population, and for this case we know the distribution for X is given by:

X \sim N(37.8,0.12)  

Where \mu=37.8 and \sigma=0.12

We are interested on this probability

P(37.5

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(37.5

And we can find this probability with this difference:

P(-2.5

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(-2.5

So then the percentage usable are 94,6%

4 0
3 years ago
Water flows through a horizontal 60 mm diameter galvanized iron pipe at a rate of 0.02 m3/s. If the pressure drop is 135 kPa per
maksim [4K]

Answer:

pipe is old one with increased roughness

Explanation:

discharge is given as

V =\frac{Q}{A} = \frac{ 0.02}{\pi \4 \times (60\times 10^{-3})^2}

V = 7.07  m/s

from bernou;ii's theorem we have

\frac{p_1}{\gamma}  +\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma}  +\frac{V_2^2}{2g} + z_2 + h_l

as we know pipe is horizontal and with constant velocity so we have

\frac{P_1}{\gamma } + \frac{P_2 {\gamma } + \frac{flv^2}{2gD}

P_1 -P_2 = \frac{flv^2}{2gD} \times \gamma

135 \times 10^3 = \frac{f \times 10\times 7.07^2}{2\times 9.81 \times 60 \times 10^{-5}} \times 1000 \times 9.81

solving for friction factor f

f = 0.0324

fro galvanized iron pipe we have \epsilon  = 0.15 mm

\frac{\epsilon}{d} = \frac{0.15}{60} = 0.0025

reynold number is

Re =\frac{Vd}{\nu} = \frac{7.07 \times 60\times 10^{-3}}{1.12\times 10^{-6}}

Re = 378750

from moody chart

For Re = 378750 and \frac{\epsilon}{d} = 0.0025

f_{new} = 0.025

therefore new friction factor is less than old friction factoer hence pipe is not new one

now for Re = 378750 and f = 0.0324

from moody chart

we have \frac{\epsilon}{d} =0.006

\epsilon = 0.006 \times 60

\epsilon = 0.36 mm

thus pipe is old one with increased roughness

5 0
3 years ago
A gas is compressed from an initial volume of 0.42 m3 to a final volume of 0.12 m3. During the quasi-equilibrium process, the pr
Georgia [21]

Answer:

W=-52 800\ \text{J}=-52.8\ \text{kJ}

Explanation:

First I sketched the compression of the gas with the help of the given pressure change process relation. That is your pressure change due to change in volume.

To find the area underneath the curve (the same as saying to find the work done) you should integrate the given relation for pressure change:

W=\int_{0.42}^{0.12}-1200V+500dV=-52.8\ \text{kJ}

6 0
3 years ago
An 18-in.-long titanium alloy rod is subjected to a tensile load of 24,000 lb. If the allowable tensile stress is 60 ksi and the
Vilka [71]

Answer:

Required Diameter = 302.65 inches

Explanation:

We are given;

Allowable tensile stress = 60 ksi

Weight of tensile load = 24,000 lb

Elongation = 0.05 in

Original length = 18 in

We'll need to check the diameters under stress and strain.

Now, we know that the formula for stress is;

Stress = Force/Area

Thus,

Area = Force/stress

So for this stress, area required is;

A_req = 24000/60 = 4000 in²

So let's find the required diameter here.

Area = πd²/4

So, 4000 = πd²/4

(4000 x 4)/π = d²

d² = 5092.96

Required diameter here is;

d = √5092.96

d = 71.36 in

For Strain;

Formula for strain is;

Strain = stress/E

We are given E = 120 ksi

stress = P/A = 24,000/A

strain = elongation/original length = 0.05/18 = 0.00278

Thus;

0.00278 = P/(A•E)

0.00278 = 24000/(120 x A)

Making A the subject to obtain;

A = 24000/(120 x 0.00278)

A_required = 71942 in²

Area = πd²/4

So, 71942 = πd²/4

(71942 x 4)/π = d²

d² = 91599.4

Required diameter here is;

d = √91599.4

d = 302.65 in

The larger diameter is 302.65 inchesand it's therefore the required one.

3 0
4 years ago
A three-point bending test is performed on a glass specimen having a rectangular cross section of height d = 5.4 mm (0.21 in.) a
Fudgin [204]

Answer:

5.21e-2mm

Explanation:

Please see attachment

8 0
3 years ago
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