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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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For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration
Amiraneli [1.4K]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that \frac{x_1^2}{Dt}=constant where x is distance and t is time .As the temperature is constant so D will be also constant

So \frac{x_1^2}{t}=constant

then \frac{x_1^2}{t_1}=\frac{x_2^2}{t_2} we have given x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm and we have to find t_2 putting all these value in equation

\frac{2^2}{15}=\frac{6^2}{t_2}

so t_2=135\ hour

5 0
3 years ago
The 50mm diameter cylinder is made from Am 1004-T61 magnesium (E = 44.7GPa, a = 26x10^-6/°C)
BartSMP [9]

Answer:

......................

Explanation:

3 0
3 years ago
A cylindrical metal specimen having an original diameter of 12.8 mm and gauge length of 50.80 mm is pulled in tension until frac
Sedaia [141]

Answer:

%Reduction in area = 73.41%

%Reduction in elongation = 42.20%

Explanation:

Given

Original diameter = 12.8 mm

Gauge length = 50.80mm

Diameter at the point of fracture = 6.60 mm (0.260 in.)

Fractured gauge length = 72.14 mm.

%Reduction in Area is given as:

((do/2)² - (d1/2)²)/(do/2)²

Calculating percent reduction in area

do = 12.8mm, d1 = 6.6mm

So,

%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²

%RA = 0.734130859375

%RA = 73.41%

Calculating percent reduction in elongation

%Reduction in elongation is given as:

((do) - (d1))/(d1)

do = 72.14mm, d1 = 50.80mm

So,

%RA = ((72.24) - (50.80))/(50.80)

%RA = 0.422047244094488

%RA = 42.20%

3 0
3 years ago
What is a Flame Front Generator?
Inessa [10]

Answer and Explanation:

Flame Front Generator: It is a ignition system which is very useful in flaring system .In this system the air and gases are mixed together and make a combustible air gas mixture. There is a flame front region where the combustion reaction takes place , it is the region where gases as like hydrogen and air mixed with each other and form combustible gases.

7 0
3 years ago
Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.
-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

                                       r = 0.5*θ

                                       θ = 0.5*t^2              ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

                                        dr/dt = 0.5*dθ/dt

                                        d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

                                        dθ/dt = t

                                        d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

                                        θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

                                        r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

                                        tan Ψ = r / (dr/dθ) = 1 / 0.5

                                        Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

                                        a_r = d^2r / dt^2 - r * dθ/dt

                                        a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

                                        a_θ =  r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

                                        a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction:              N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction:                      F  - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

                                       N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

                                       N_c = 3.03 N

                                       F  - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

                                       F = 1.81 N

4 0
3 years ago
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