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3 years ago

15

Going back the b beginning of the process is common in engineering true or false?

1 answer:

Yuliya22 [10]3 years ago

6 0

Answer:

true !!

Explanation:

hope i helped :)! brainliest ?

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Does anyone know the answer to this

soldi70 [24.7K]

Answer:

A

Explanation:

7 0

2 years ago

Read 2 more answers

Architecture reflects multidisciplinary

vovangra [49]

Answer:

The architectural profession is multidisciplinary. ... Some of the disciplines which intervene in the design of a building are: architectural composition, structural engineering, architectural technology and energy engineering. Depending on the strategy of designing, one or more disciplines can have priority over others..

Explanation:

Hope it helps you..

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6 0

3 years ago

In a tensile test on a steel specimen, true strain is 0.171 at a stress of 263.8 MPa. When true stress is 346.2 MPa, true strain

swat32

Answer:n=0.973

Explanation:

Given

When True strain $\left ( \epsilon _T_1\right )=0.171$

at $\sigma _1=263.8 MPa$

When True stress $\left ( \sigma _2\right )$ =346.2 MPa

true strain $\left ( \epsilon _T_2\right )$ =0.226

We know

$\sigma =k\epsilon ^n$

where $\sigma$ =True stress

$\epsilon$ =true strain

n=strain hardening exponent

k=constant

Substituting value

$263.8=k\left ( 0.171\right )^n------1$

$346.2=k\left ( 0.226\right )^n-----2$

Divide 1 & 2 to get

$\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n$

$1.312=\left ( 1.3216\right )^n$

Taking Log both side

$ln\left ( 1.312\right )=nln\left ( 1.3216\right )$

n=0.973

6 0

3 years ago

Two Carnot engines operate in series such that the heat rejected from one is the heat input to the other. The heat transfer from

kykrilka [37]

Answer:

Given:

high temperature reservoir $T_{H} =1000k$

low temperature reservoir $T_{L} =400k$

thermal efficiency $n_{1}= n_{2}$

The engines are said to operate on Carnot cycle which is totally reversible.

To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as

$n_{1} =1-\frac{T}{T_{H} }$

The thermal efficiency of second heat engine can be written as

$n_{2} =1-\frac{T_{L} }{T}$

The temperature of intermediate reservoir can be defined as

$1-\frac{T}{T_{H} } =1-\frac{T_{L} }{T} \\T^2=T_{L} T_{H} \\T=\sqrt{T_{L} T_{H} }\\T=\sqrt{400*1000} =632k$

8 0

3 years ago

The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular veloc

madreJ [45]

Answer:

Option B

$116 ft/s^{2}$

Explanation:

$\theta=2 rev=2(2\pi)=4\pi$

$\alpha \theta=0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (4\pi)= 0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-4^{2})$

$\omega_f=8.14 rads/s$

$v=r\omega=1.75*8.14=14.245 ft/s$

Centripetal acceleration $=\omega_f^{2} r=8.14^{2}*1.75=115.95 ft/s^{2}$

Tangential component=dr=2*1.75=3.5

$Resultant=\sqrt{3.5^{2}+115.95^{2}}\approx 116 ft/s^{2}$

5 0

3 years ago