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4 years ago

Any programmer who writes a Diophantine equation solver must occasionally encounter an infinite loop.

Engineering

1 answer:

Sergio [31]4 years ago

4 0

Answer: True

Explanation:

An Infinite loop occurs in a computer programming when a sequence of instructions run endlessly without stopping until there is an external intervention, this intervention could be pull or plug. It is also known as endless loop. It occurs due to using of variables that are not properly updated or when there is an error in looping condition.

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For each function , sketch the Bode asymptotic magnitude and asymptotic phase plots.

horrorfan [7]

Answer:

attached below

Explanation:

a) G(s) = 1 / s( s+2)(s + 4 )

Bode asymptotic magnitude and asymptotic phase plots

attached below

b) G(s) = (s+5)/(s+2)(s+4)

phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4

attached below

c) G(s)= (s+3)(s+5)/s(s+2)(s+4)

solution attached below

5 0

3 years ago

At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =

Ahat [919]

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

$F=ma$ for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that $W = mg$

For simplicity we work with $g =9.807$ $\frac{m}{s^{2}}$ despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula $g = a-bz$ the "normal" or "standard" weight of an object is given by $W = mg = ma$ when $z = 0$ , so we need to find the value of $z$ that makes $W = m(a-bz) = 0.997ma$ meaning that the original weight decrease by a 0.30%, so now we operate...

$m(a-bz) = 0.997ma$ now we group like terms on the same sides $ma(1-0.997) = mbz$ we cancel equal tems on both sides and obtain that $z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m$

7 0

3 years ago

There is evidence that liquid water existed on Mars at some time in the past. What is the evidence found indicating the presence

Norma-Jean [14]

There mag be water on Mars but it’s not very easy to spot, but that is besides the point,

Mars is a smaller planet then Earth, thus it has less gravity, meaning when water evaporates into the atmosphere of the planet the water slowly escapes into space, so there is less and less water on the planet.

Hope this helps!

5 0

3 years ago

A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe

vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V $_f$ = V $_g$ = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v $_f$ = 0.001057 m³/kg

v $_g$ = 1.0037 m³/kg

u $_f$ = 486.82 kJ/kg

u $_g$ 2524.5 kJ/kg

h $_g$ = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V $_f$ /v $_f$ + V $_g$ /v $_g$

we substitute

m₁ = 0.002/0.001057 + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v $_g$

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m $_{out$ = m₁ - m₂

m $_{out$ = 1.89414 - 0.003985

m $_{out$ = 1.890155 kg

so, Initial internal energy will be;

U₁ = m $_f$ u $_f$ + m $_g$ u $_g$

U₁ = (V $_f$ /v $_f$ )u $_f$ + (V $_g$ /v $_g$ )u $_g$

we substitute

U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E $_{in$ - E $_{out$ = ΔE $_{sys$

QΔt - m $_{out$ h $_{out$ = m₂u₂ - U₁

QΔt - m $_{out$ h $_g$ = m₂u $_g$ - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0

3 years ago

The way most recursive functions are written, they seem to be circular at first glance, defining the solution of a problem in te

EastWind [94]

Question Continuation

int factorial(int n) {

if(n == 0)

return 1;

else

return n * factorial(n - 1);

}

Provide a brief explanation why this recursive function works.

Show all steps involved in calculating factorial(3) using the function deﬁned.

Answer:

1. Brief explanation why this recursive function works.

First, the recursive method factorial is defined.

This is the means through with the machine identifies the method.

The method is defined as integer, the machine will regard it as integer.

When the factorial is called from anywhere that has access to it, which in this case is within the factorial class itself. This means you can call it from the main method, or you can call it from the factorial method itself. It's just a function call that, well, happens to call itself.

2. Steps to calculate factorial(3)

1 First, 3 is assigned to n.

2. At line 2, the machine checks if n equals 0

3. If yes, the machine prints 1

4. Else; it does the following from bottom to top

factorial(3):

return 3*factorial(2);

return 2*factorial(1):

return 1;

Which gives 3 * 2 * 1 = 6

5. Then it prints 6, which is the result of 3!

6 0

3 years ago