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2 years ago

Please look at the picture for question.

Engineering

2 answers:

igor_vitrenko [27]2 years ago

6 0

Answer:

more than10 joules

Explanation:

Vilka [71]2 years ago

5 0

Answer:

more than 10 joules jjsjdnjdfjnfjfnjfnnfndnndjnfjgngjntjfjmrmdkdmjd

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An existing building is suffering from cracks in the exterior walls. The investigating engineer wants to ensure that the foundat

jek_recluse [69]

Answer:

$18 ft^{2}$

Explanation:

Soil bearing pressure= $\frac {Load}{Area}$

Since we're given pressure of 2500 psf and load of 45000 pounds

The area= $\frac {45000}{2500}=18$

Therefore, the smallest area of safe footings should not be less than $18 ft^{2}$

6 0

3 years ago

g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th

xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

$qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0$

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }$

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }$

if Gl≈El(l,5800)

$\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt= 2*5800=11600 um-K, at this value, F=0.941

$\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77$

The hemispherical emissivity is equal to:

$E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt=2*333=666 K, at this value, F=0

$E=0+(1-0.7)(1)=0.3$

The hemispherical absorptivity is equal to:

$qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}$

3 0

3 years ago

A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 54 percent. Determine the rate of hea

Gennadij [26K]

Answer:

$\dot Q _{L} = 511.111 MW$ . Heat transfer can be higher if themal efficiency is lower.

Explanation:

The heat transfer rate to the river water is calculated by this expression:

$\dot Q_{L} = \dot Q_{H} - \dot W$

$\dot Q_{L} = (\frac{1}{\eta_{th}}-1 )\cdot \dot W\\\dot Q_{L} = (\frac{1}{0.54}-1)\cdot (600 MW)\\\dot Q _{L} = 511.111 MW$

The actual heat transfer can be higher if the steam power plant reports an thermal efficiency lower than expected.

8 0

4 years ago

In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o

lesya [120]

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

= $10\times 101325 \ Pa$

= $1013250 \ Pa$

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

= $1 \ m^3$

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ $PV=nRT$

⇒ $n=\frac{PV}{RT}$

By substituting the values, we get

$=\frac{1013250\times 1}{8.3145\times 298}$

$=408.94 \ moles$

As we know,

⇒ $Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}$

or,

⇒ $MW=\frac{m}{n}$

$=\frac{11.5}{408.94}$

$=0.02812 \ Kg/mol$

$=28.12 \ g/mol$

3 0

3 years ago

Say you have a random, unordered list containing 4096 four-digit numbers. Describe the most efficient way to: sort the list and

Debora [2.8K]

Answer:

Answer explained below

Explanation:

It is given that numbers are four-digit so maximum value of a number in this list could be 9999.

So we need to sort a list of integers, where each integer lies between [0,9999].

For these given constraints we can use counting sort which will run in linear time i.e. O(n).

--------------------------------------------------------------------------------

Psuedo Code:

countSort(int numList[]) {

int count[10000];

count[i] = 0; for all i;

for(int num in numList){

count[num]+= 1;

}

return count;

}

--------------------------------------------------------------------------------

Searching in this count array will be just O(1).

E.g. Lets say we want to search if 3 was present in the original list.

Case 1: it was present in the original list:

Then the count[3] would have been incremented by our sorting algorithm. so in case element exists then count value of that element will be greater than 0.

Case 2: it was not present:

In this case count[3] will remain at 0. so in case element does not exist then count of that element will be 0.

So to search for an element, say x, we just need to check if count[x]>0.

So search is O(1).

Run times:

Sorting: O(n)

Search: O(1)

6 0

3 years ago