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3 years ago

Driving Distraction Brainstorming Session

1 answer:

5 0

texting, phone calls, putting on makeup, brushing hair, movies playing in car, loud music, children, and that's pretty much all I could think of

please give <u>BRAINLIEST ANSWER └[T‸T]┘</u>

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The ratio of boys to girls in the class was 2 to 3. If there were 18 girls in the class, how many students were in the class

iren2701 [21]

37 students in the class
18 Girls and 19 Boys

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3 years ago

QUICK ASAP!!!

nignag [31]

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Im pretty sure the answer is B

Explanation:

Paul will need to persuade his client of the benefits of the different filter.

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2 years ago

Para uma dada velocidade do vento, o gerador de turbina mostrado na figura produz 22 kW de potência elétrica. Como um sistema co

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Answer:

Explanation:

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3 years ago

The conditions at the beginning of compression in an Otto engine operating on hot-air standard with k=1.35 and 101.325 kPa, 0.05

Katyanochek1 [597]

Answer:

$P_m=181.42 KPa$

Explanation:

$P_1=101.325 KPa,V_1=0.05m^3,T_1=32C$ ,K=1.35

Clearance is 8%.

Heat added=15 KJ

We know that compression ratio $r=1+\dfrac{1}{C}$

$r=1+\dfrac{1}{0.08}$

r=13.5

$r=\dfrac{V_1}{V_2}$

$13.5=\dfrac{0.05}{V_2}$

$V_2=3.7\times 10^{-3}$

We know that efficiency of otto cycle

$\eta =1-\dfrac{1}{r^{k-1}}$

$\eta =1-\dfrac{1}{13.5^{1.35-1}}$

$\eta =0.56$

$\eta =\dfrac{W}{Q}$

W is the work out put and Q is the heat addition.

$0.56 =\dfrac{W}{15}$

W=8.4 KJ

We know that Work =Mean effective pressure x swept volume.

Here swept volume $V_s=V_1-V_2$

$V_s=0.05-3.7\times 10^{-3}$

$V_s=0.0463 m^3$

Noe by putting the values

Work =Mean effective pressure x swept volume.

$8.4=P_m\times 0.0463$

$P_m=181.42 KPa$

6 0

3 years ago

Tungsten is being used at half its melting point (Tm≈3,400◦C) and astress level of 160 MPa. An engineer suggests increasing the

Paladinen [302]

Answer:

Explanation:

The missing diagram is attached in the image below which shows the deformation map of the Tungsten.

Given that:

Stress level $\sigma = 160 MPa$

T = 0.5 Tm

$\implies \dfrac{T}{Tm} = 0.5$

G = 160 GPa

$\implies \dfrac{\sigma}{G} = 10^{-3}$

The regulating creep mechanism is dislocation driven, as we can see from the deformation mechanism.

The engineer's recommendation would not be approved because increasing grain size results in a decrease in the grain-boundary count, preferring dislocation motion. The existence of grain borders is a hindrance to dislocation motion, as the dislocation principle explicitly states. To stop the motion, we'll need a substance with finer grains, which would result in more grain borders, or a material with higher pressure. In the case of Nabarro creep, which is diffusion-driven, an engineer's recommendation would be useful.

If stress level reduced to $\sigma = 1.6 MPa$

$\implies \dfrac{\sigma }{G} = 10^{-5}$

Cable creep is now the controlling creep mode, which entails tension-driven atom diffusion along grain borders to elongate grain along the stress axis, a process known as grain-boundary diffusion. Cable creep is more common in fine-grained materials. As a result, the engineer's advice would succeed in this case. The affinity for cable creep is reduced when the grain size is increased.

From the map of creep mechanism for $\dfrac{\sigma}{G} = 10^{-3} \ and \ \dfrac{T}{Tm} = 0.5$