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3 years ago

CUse the drop-down menus to complete the statements.

Engineering

2 answers:

Illusion [34]3 years ago

7 0

Answer:

The expansion of the Roman Empire helped grow trade.

Caesar Augustus stabilized prices and promoted trade by establishing a common currency.

Roads helped the economy by connecting inland areas to water routes.

Improvements in agriculture led to better trade and made Rome less dependent on wheat from Egypt.

Iteru [2.4K]3 years ago

4 0

Elmo World lalalalalalalalaalalalalalalalalalalalalalalalalalalaal it’s Elmo World it’s Elmo World
Explanation

See eLmO WoRlD

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What is the stress concentration factor of a shaft in torsion, where D=1.25 in. and d=1 in. and the fillet radius is, r=0.2 in.a

klio [65]

Answer:

Concentration factor will be 1.2

So option (C) will be correct answer

Explanation:

We have given outer diameter D = 1.25 in

And inner diameter d = 1 in and fillet ratio r = 0.2 in

So $\frac{r}{d}$ ratio will be $=\frac{0.2}{1}=0.2$

And $\frac{D}{d}$ ratio will be $=\frac{1.25}{1}=1.25$

Now from the graph in shaft vs torsion the value of concentration factor will be 1.2

So concentration factor will be 1.2

So option (C) will be correct answer.

6 0

3 years ago

When was solar power envold

Lyrx [107]

Answer:

1839

Explanation:

the first huge step in solar penal technology came when alxendare becquerel observed the photovolatic effect in 1839,which occurs when a material produces electric current when exposed to light.it was not untill 1888 that the first solar cell was actually built by aleksander stoletov.

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3 years ago

Read 2 more answers

A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a

nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, $g_{c}$ = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

$c_{p}$ = 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\$

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence $v_{a} ^{2} = 0$

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }$