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4 years ago

5

Given the strings s1 and s2, not necessarily of the same length, create a new string consisting of alternating characters of s1

and s2 (that is, the first character of s1 followed by the first character of s2, followed by the second character of s1, followed by the second character of s2, and so on. Once the end of either string is reached, no additional characters are added. For example, if s1 contained "abc" and s2 contained "uvwxyz", then the new string should contain "aubvcw". Assign the new string to the variable s3.

1 answer:

andrew11 [14]4 years ago

4 0

Answer:

THE ANSA IZ A

Explanation:

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2.24 Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same s

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Explanation:

Given that:

From process 1 → 2

$P_1 = 10 bar \\ \\ V_1 = 1 m^3 \\ \\ V_2 = 4 m^3$

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$\gamma = 1.5$

Process 2 → 3

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$P_3 = 10 \ bar$

Process 3 → 1

P = constant i.e the compression from state 1

Now, to start with 1 → 2

$P_1V_1^{1.5} = P_2V_2^{1.5}$

$P_2 = P_1 (\dfrac{V_1}{V_2})^{1.5}$

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The work-done for the process 1 → 2 through adiabatic expansion is:

$W = \dfrac{1}{1-\gamma}[P_2V_2-P_1V_1]$

We know that 1 bar = $10^5 \ N/m^2$

∴

$W = \dfrac{1}{1-1.5}[1.25 \times 10^5 \times 4- 10 \times 10^5 \times 1]$

$W =1000000 \ J$

$W_{1 \to 2} = 1000 kJ$

For process 2 → 3

Since V is constant

Thus:

W = PΔV = 0

$W_{2 \to 3} = 0$

For process 3 → 1

W = PΔV

$W _{3 \to 1} = P_3(V_1-V_3)$

$W _{3 \to 1} = 10 \times 10^5 (1-4)$

$W _{3 \to 1} = 10 \times 10^5 (-3)$

$W _{3 \to 1} = -3 \times 10^6 \ J$

$W _{3 \to 1} = -3000 \ kJ$

The net work-done now for the entire system is :

$W_{net} = W_{1 \to 2} + W_{2 \to 3 } + W_{ 3 \to 1 }$

$W_{net} = (1000 + 0 + (-3000)) \ kJ$

$W_{net} =-2000 \ kJ$

The sketch of the processes on p -V coordinates can be found in the image attached below.

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