4 years ago

What is NOT an example of professional networking?

Engineering

1 answer:

gogolik [260]4 years ago

6 0

Answer:

funfairs etc

Explanation:

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What is the minimum efficiency of a functioning current-model catalytic converter? a. 60% b. 75% c. 80% d. 90%

slamgirl [31]

Answer:

d. 90%

Explanation:

As we know that internal combustion engine produce lot's of toxic gases to reduce these toxic gases in the environment a device is used and this device is know as current modeling converter.

Generally the efficiency of current model catalytic converter is more than 90%.But the minimum efficiency this converter is 90%.

So option d is correct.

d. 90%

7 0

3 years ago

I accidently peed my pants help me change me pls

nataly862011 [7]

Answer: *changed*

Explanation: Because you peed

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3 years ago

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Rect answer in the box. Spell all Words correctly

inna [77]

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3 years ago

A long rod of 60-mm diameter and thermophysical properties rho= 8000 kg/m3, c= 500 J/kg·K, and k= 50 W/m·K is initially at a uni

Dvinal [7]

Answer:

Tc = = 424.85 K

Explanation:

Data given:

D = 60 mm = 0.06 m

$\rho = 8000 kg/m^3$

k = 50 w/m . k

c = 500 j/kg.k

$h_{\infty} = 1000 w/m^2$

$t_{\infity} = 750 k$

$t_w = 500 K$

$surface area = As = \pi dL$

$\frac{As}{L} = \pi D = \pi \timeS 0.06$

HEAT FLOW Q is

$Q = h_{\infty} As (T_[\infty} - Tw)$

$= 1000 \pi\times 0.06 (750-500)$

= 47123.88 w per unit length of rod

volumetric heat rate

$q = \frac{Q}{LAs}$

$= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}$

$q = 1.66\times 10^{7} w/m^3$

$Tc = \frac{- qR^2}{4K} + Tw$

$= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 56} + 500$

= 424.85 K

7 0

4 years ago

Calculate the maximum load that a 7075 series aluminum alloy bar (with a T6 temper heat treatment) can support without permanent

Aleksandr [31]

Answer:

The maximum load the bar can withstand = 35.43 KN

Explanation:

Ultimate tensile strength of the given aluminium bar $\sigma$ = 540 M pa

Cross section area of the bar = $8.1^{2}$ = 65.61 $mm^{2}$

We know that the ultimate strength of the bar is calculated from

$\sigma = \frac{P_{max} }{A}$

$540 = \frac{P_{max} }{65.61}$

$P_{max}$ = 540 × 65.61

$P_{max}$ = 35.43 KN

Therefore the maximum load the bar can withstand = 35.43 KN

6 0

4 years ago