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levacccp [35]
3 years ago
14

The steel 4140 steel contains 0.4% C, however, it shows higher yield strength and ultimate strength than that of the 1045 (0.45%

C explain why)
Engineering
1 answer:
Aleonysh [2.5K]3 years ago
4 0

Answer:

4140 steel contains 0.4% C  having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

Explanation:

we have given 4140 steel contains 0.4% C

we know here that 4140 steel is low steel alloy , and it have low amount of chromium , manganese etc alloying element

and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel

while in 1045 steel contains 0.45 % c is plain carbon steel

and it do not contain any alloying element

so that 4140 steel contains 0.4% C  having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

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leonid [27]

Answer:

heat transfer for the process is - 643.3 kJ

Explanation:

given data

mass m = 2 kg

pressure p1 = 500 kPa

temperature t1 = 400°C = 673.15 K

temperature t2 = 40°C = 313.15 K

pressure p2 = 300 kPa

to find out

heat transfer for the process

solution

we know here mass is constant so

m1 = m2

so by energy equation

m ( u2 - u1 ) = Q - W

Q is heat transfer

and in process  P =  A+ N that is linear spring

so

W = ∫PdV

= 0.5 ( P1+P2) ( V1 - V2)

so for case 1

P1V1 = mRT

put here value

500 V1 = 2 (0.18892) (673.15)

V1 = 0.5087 m³

and

for case 2

P2V2 = nRT

300 V2 = 2 (0.18892) (313.15)

V2 = 0.3944 m³

and

here W will be

W = 0.5 ( 500 + 300 ) ( 0.3944 - 0.5087 )

W = -45.72 kJ

and

Q is here for Cv = 0.83 from ideal gas table

Q = mCv ( T2-T1 ) + W

Q = 2 × 0.83 ( 40 - 400 ) - 45.72

Q = - 643.3 kJ

heat transfer for the process is - 643.3 kJ

7 0
3 years ago
Examine a process whereby air at 300 K, 100 kPa is compressed in a piston/cylinder arrangement to 600 kPa. Assume the process is
professor190 [17]

Answer:

See attachment and explanation.

Explanation:

- The following question can be solved better with the help of a MATLAB program as follows. The code is given in the attachment.

- The plot of the graph is given in attachment.

- The code covers the entire spectrum of the poly-tropic range ( 1.2 - 1.6 ) and 20 steps ( cases ) have been plotted and compared in the attached plot.

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Why is tubing sometimes coiled when installed in a car or vechile
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Answer:

Coiled tubing is often used to carry out operations similar to wire lining.

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yan [13]

Answer:

b. 1232.08 km/hr

c. 1.02 kn

Explanation:

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See attachment for the remaining.

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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).

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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:

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Search: O(1)

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3 years ago
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