What is the value of the magicPowers variable after executing the following code snippet? String magicPowers = ""; int experienc
eLevel = 9; if (experienceLevel > 10) { magicPowers = magicPowers + "Golden sword "; } if (experienceLevel > 8) { magicPowers = magicPowers + "Shining lantern "; } if (experienceLevel > 2) { magicPowers = magicPowers + "Magic beans "; }
1 answer:
Answer:
Shining lantern Magic beans
Explanation:
I will explain the code line by line.
The first statement contains a string type variable magicPowers and second statement has an int type variable experienceLevel which is initialized by value 9.
Nested if statement are used and lets see which of those statements execute.
The first if statement checks if the value of experienceLevel is greater than 10. This statement evaluates to false because the value of experienceLevel is set to 9. Then the program control moves to the next if statement.
The second if statement checks if the value of experienceLevel is greater than 8. This if condition evaluates to true because the value of experienceLevel is 9. So this if statement is executed which contains the statement magicPowers = magicPowers + "Shining lantern "; . This means magicPowers stores the string Shining lantern.
The third if statement checks if the value of experienceLevel is greater than 2. As the value of experienceLevel is 9 so this condition also evaluates to true which means the statement magicPowers = magicPowers + "Magic beans "; is executed. This statement means magicPowers stores the string Magic Beans
You can use a print statement to display the output on the screen as:
System.out.print(magicPowers);
Output:
Shining lantern Magic beans
The program along with the output is attached as a screenshot.
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Answer:
the ratio of the etched to the original crack tip radius is 30.24
Explanation:
Given the data in the question;
we determine the initial fracture stress using the following expression;
(σf)₁ = 2(σ₀)₁ α₁/(
)₁
----- let this be equation 1
where; (σ₀)₁ is the initial fracture strength
()₁ is the original crack tip radius
α₁ is the original crack length.
first, we determine the final crack length;
α₂ = α₁ - 16% of α₁
α₂ = α₁ - ( 0.16 × α₁)
α₂ = α₁ - 0.16α₁
α₂ = 0.84α₁
next, we calculate the final fracture stress;
the fracture strength is increased by a factor of 6;
(σ₀)₂ = 6( σ₀ )₁
Now, expression for the final fracture stress
(σf)₂ = 2(σ₀)₂ α₂/(
)₂
------- let this be equation 2
where ()₂ is the etched crack tip radius
value of fracture stress of glass is constant
Now, we substitute 2(σ₀)₁ α₁/(
)₁
from equation for (σf)₂ in equation 2.
0.84α₁ for α₂.
6( σ₀ )₁ for (σ₀)₂.
∴
2(σ₀)₁ α₁/(
)₁
= 2(6( σ₀ )₁)
0.84α₁/(
)₂
divide both sides by 2(σ₀)₁
α₁/(
)₁
= 6
0.84α₁/(
)₂
1/(
)₁
= 6
0.84/(
)₂
1/(
)₁
= 36
0.84/(
)₂
1 / ()₁ = 30.24 / (
)₂
()₂ = 30.24(
)₁
()₂/(
)₁ = 30.24
Therefore, the ratio of the etched to the original crack tip radius is 30.24
Answer: Pi= 4 - 4/3 + 4/5 - 4/7 + 4/9 ...
Explanation:
Is the same as the example,
If Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...
Then
(Π/4 )*4= 4*(1 - 1/3 + 1/5 - 1/7 + 1/9 ...)
Π =4 - 4/3 + 4/5 - 4/7 + 4/9 ...
The way to write this is
Sum(from n=0 to n=inf) of (-1)^n 4/(2n+1)
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Given :
Force, .
Force is acting at point A( 2 m, 3 m ) and B( 3 m, 5 m )
To Find :
The work done by force F .
Solution :
Displacement vector between point A and B is :
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W = 12000 J
Therefore, work done by force is 12000 J .