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boyakko [2]
3 years ago
5

Your task is to fill in the missing parts of the C code to get a program equivalent to the generated assembly code. Recall that

the result of the function is returned in register %eax. You will find it helpful to examine the assembly code before, during, and after the loop to form a consistent mapping between the registers and the program variables.
Engineering
1 answer:
Rudik [331]3 years ago
6 0

Answer:

See Explaination

Explanation:

//Function

long loop (long x, long n)

{

//Declare a variable named result and initialize it to zero

long result = 0;

//Declare a variable named mask

long mask;

//For loop

for(mask = 1; mask != 0; mask = mask << (n & 0xFF))

{

//Calculate

result | = (x&mask);

}

//Return result

return result;

}

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Brake system

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Valorant Or Csgo? which is best in your opinion?
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1) Relative to electrons and electron states, what does each of the four quantum numbers specify?
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Cite two important additional refinements that resulted from the Wave-mechanical

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3 years ago
Very thin films are usually deposited under vacuum conditions to prevent contamination and ensure that atoms can fly directly fr
katrin [286]

Answer:

a. 9947 m

b. 99476 times

c. 2*10^11 molecules

Explanation:

a) To find the mean free path of the air molecules you use the following formula:

\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

n_{collision}=\frac{9947.62m}{0.05m}\approx198952\  times

c) By using the equation of the ideal gases you obtain:

PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules

5 0
3 years ago
If a front gear had 24 teeth, and a rear gear has 12 teeth:
zubka84 [21]

Answer:

  4 times around

Explanation:

The total number of teeth involved will be the same for each gear. If the front gear is connected to the pedal and it goes around twice, then 2·24 = 48 teeth will have passed the reference point.

If the rear gear is attached to the wheel, and 48 teeth pass the reference point, then it will have made ...

  (48 teeth)/(12 teeth/turn) = 4 turns

4 0
3 years ago
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