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Rudiy27
3 years ago
9

What will the following segment of code output? score = 95; if (score > 95) cout << "Congratulations!\n"; cout <<

"That's a high score!\n"; cout << "This is a test question!" << endl; Answers: That's a high score! This is a test question! Congratulations! That's a high score! This is a test question! This is a test question! Congratulations! That's a high score! None of these
Engineering
1 answer:
Anarel [89]3 years ago
5 0

Answer:

That's a high score!

This is a test question!

Explanation:

The reason these two lines are printed and not the first one is simple. After the 'IF' condition has been stated, there is no use of parenthesis such as { and } to enclose the next lines. This means that only the first line after the 'IF' condition may be read or skipped depending on whether the condition (score>95) is met. Since the score is not larger than 95, and the 'IF' condition fails, the line 'Congratulations!' is not printed. The next two lines of the code are read as normal because they do not depend on the 'IF' condition.

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Which goal incorporates most of the criteria required for a SMART goal?
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Explanation:

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3 years ago
A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the
raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

E = G \times H

Where G represents complex rated power and H is the inertia constant of turbo-generator.

E = 100 \times 8 \\\\E = 800 \: MJ

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$ P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}  $

Where M is given by

$ M = \frac{E}{180 \times f} $

$ M = \frac{800}{180 \times 50} $

M = 0.0889 \: MJ \cdot s/ elec \: \: deg

So, the rotor acceleration is

$ P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}  $

$  30 = 0.0889 \frac{d^2 \delta}{dt^2}  $

$   \frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}  $

$   \frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2 $

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$ \Delta  \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2 $

Where t is given by

1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2  \\\\t = 0.2 \: sec

So,

$ \Delta  \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2 $

$ \Delta  \delta = 6.75 \: elec \: deg

The change in torque in rpm/s is given by

$ \Delta  \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ  }   $

$ \Delta  \delta =28.12 \: \: rpm/s $

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$ Rotor \: speed = \frac{120 \cdot f}{P}  + (\Delta  \delta)\cdot t  $

Where P is the number of poles of the turbo-generator.

$ Rotor \: speed = \frac{120 \cdot 50}{4}  + (28.12)\cdot 0.2  $

$ Rotor \: speed = 1500  + 5.62  $

$ Rotor \: speed = 1505.62 \:\: rpm

4 0
3 years ago
Let r be a bank's interest rate in percent per year (APR). An initial amount of money P, also called as principal, will mature t
Liono4ka [1.6K]

Answer:

See explanation below.

Explanation:

For this case the program needs to take the inputs as P,r and n and the output would be as A and printed on the system. The code is:

# Inputs

P = float(input("Enter the present value : "))  

r = float(input("Enter your APR : "))  

n = float(input("Enter the number of years : ") )

# Output

A = P*(1 +(r/100))**n

print("The future values is:", A)  

And the result obtained is:

Enter the present value : 1000

Enter your APR : 0.95

Enter the number of years : 5

The future values is: 1048.4111145526908

7 0
3 years ago
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