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3 years ago

If i see i dead army person will i be asked questions and if i see the person who did it

Engineering

2 answers:

True [87]3 years ago

5 0

Answer:

Yes we take dead army people like it was on purspose because people would like to kill us off but yes you would be asked Questions and we would ask you if you saw any suspects

evablogger [386]3 years ago

5 0

Answer:

yes true it will be

Explanation:

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What is another term for the notes that a reader can add to text in a word-processing document?

melamori03 [73]

Answer:

annotations maybe?

4 0

3 years ago

Read 2 more answers

For the same cross-sectional area, which column provides the higher buckling load: a circular bar or a circular tube?

juin [17]

Answer:

Circular tube

Explanation:

Now for better understanding lets take an example

Lets take

Diameter of solid bar= $4\sqrt{2}$ cm

Outer diameter of tube =6 cm

Inner diameter of tube=2 cm

So from we can say that both tubes have equal cross sectional area.

We know that buckling load is given as $P = \dfrac{\pi ^2EI}{L_e^2}$

If area moment of inertia(I) is high then buckling load will be high.

We know that area moment of inertia(I)

For circular tube $I = \dfrac{\pi }{64}(D_o^4-D_i^4)$

For circular bar $I = \dfrac{\pi }{64}D^4$

Now by putting the values

For circular tube $I=62.83 cm^4$

For circular bar $I=50.26 cm^4$

So we can say that for same cross sectional area the area moment of inertia(I) is high for tube as compare to bar.So buckling load will be higher in tube as compare to bar.

3 0

3 years ago

The following median grain size data were obtained during isothermal liquid phase sintering of an 82W-8Mo-8Ni-2Fe alloy. What is

Morgarella [4.7K]

Answer:

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

( $d^{2}$ - $d_{0} ^{2}$ = kt )

Explanation:

The plot attached below shows the time dependence of the growth of grain.

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

the ideal growth follows this principle = $d^{2} - d^{2} _{0}$ = kt

d = final grain size

$d_{0}$ = initial grain size

k = constant ( temperature dependent )

t = 0

8 0

3 years ago

Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l

sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is $0.44 \frac{W}{m^{2} \times K}$ , the average convection heat transfer coefficient over the entire plate is $0.293 \frac{W}{m^{2} \times K}$ and the total heat flux transfer to the plate is $61.6 KJ$ .

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas and surface. Further temperature boundary layer will developed on flat plate in longitudinal direction.

Hot carbon dioxide exhaust gas

physical properties

$r= 1.05 \frac{kg}{m^{3}}$

$c_p = 1.02 \frac{kJ}{Kg \times K}$

$m= 231 \times 10^{7} \frac{N \times s }{m^2}$

υ = $21.8 \times 10^{6} \frac{m^2}{s}$

$k = 32.5 \times 10^{3} \frac{W}{m \times K}$

$\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}$

$Pr = 0.725$

Apart from these other data arr given below,

$v= 3 \frac{m}{s} \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C$

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

$Nu = \frac{ h \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

where h= Average heat transfer coefficient

L= Length of a plate

k= Thermal Conductivity of carbon dioxide

Re = Reynold's Number

Pr = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

is referred as local convection heat transfer coefficient.

To find convection heat transfer coefficient at 1 m from leading edge,

$Nu = \frac{ h_local \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

Here, first we have to find Re and Pr,

$Re = \frac{r \times v \times L}{m}$

$Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}$

$Re = 20.63 \times 10^{-10}$

Pr number is take from physical property data and Pr is 0.725.

Putting value of Re and Pr in main equation,

we get

$Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 32.5 \times 10^{3} \times 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 0.44 \frac{W}{m^{2} \times K}$

(b) To find average convection heat transfer coefficient,

it can be find out as case (a), only difference is that instead of L=1 m, L=1.5 m would come,

Therefore,

$Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

Finally,

$h = \frac{0.44}{1.5}$

$h = 0.293 \frac{W}{m^{2} \times K}$

$Q = h \times (T_g - T_s)$

$Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140 \\ Q= 61.6 KJ$

8 0

3 years ago

What entrepreneurial activities do you know?are you capable of doing entrepreneurial activities

kherson [118]

Answer:

.,m

Explanation:

.,m

5 0

3 years ago