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maksim [4K]
3 years ago
8

I’m bored let’s text

Engineering
2 answers:
aleksandrvk [35]3 years ago
7 0

Answer:

anyways thnx for the points and have great day

Nuetrik [128]3 years ago
3 0

Answer:

<em>hi</em><em> </em><em>sa</em><em>m</em><em>e</em><em> </em><em>here</em><em>.</em><em> </em><em>hru</em><em> </em><em>btw</em>

<em>안녕하세요 모두가 잘되기를 바랍니다. 안전 유지</em>

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Sarah needs to create an architectural drawing for a museum building with an inclined surface. Which presentation view will be t
prohojiy [21]

Answer: auxiliary

Explanation: got it right

7 0
2 years ago
You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the
Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, \sigma /Strain, ∈

initial Strain, \epsilon_i = \frac{\sigma}{E}

\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}

\epsilon_i = 0.002

creep rate in the steady state

\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )

\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )

but Tinitial = 0

\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001

\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0
3 years ago
In a certain chemical plant, a closed tank contains ethyl alcohol to a depth of 71 ft. Air at a pressure of 17 psi fills the gap
Yuliya22 [10]

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

5 0
3 years ago
Engineering controls are the physical changes that employers make to the work environment or to equipment that make it safer to
____ [38]

Answer:

Engineering Controls. The best engineering controls to prevent heat-related illness is to make the work environment cooler and to reduce manual workload with mechanization. A variety of engineering controls can reduce workers' exposure to heat: Air conditioning, Increased general ventilation , Cooling fans , Local exhaust ventilation at points of high heat production or moisture, Reflective shields to redirect radiant heat , Insulation of hot surfaces Elimination of steam leaks , Cooled seats or benches for rest breaks , Use of mechanical equipment to reduce manual work, Misting fans that produce a spray of fine water droplets.

Hope this helped you!

Explanation:

5 0
3 years ago
What happens to the odometer reading when a car drives beyond its maximum reading?
Olin [163]
The odometer keeps running after you move beyond its upper limit, but the largest place values cannot be displayed due to overflow error.
8 0
2 years ago
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