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posledela
3 years ago
13

Write a program to calculate overtime pay of 10 employees. Overtime is paid at the rate of Rs. 12.00

Engineering
1 answer:
fgiga [73]3 years ago
6 0

Answer:

Here is the code.

Explanation:

#include<stdio.h>

int main()

{

int i, time_worked, over_time, overtime_pay = 0;

for (i = 1; i <= 10; i++)

{

 printf("\nEnter the time employee worked in hr ");

 scanf("%d", &time_worked);

 if (time_worked>40)

 {

  over_time = time_worked - 40;

  overtime_pay = overtime_pay + (12 * over_time);

 }

}

printf("\nTotal Overtime Pay Of 10 Employees Is %d", overtime_pay);

return 0;

}

Output :

Enter the time employee worked in hr 42

Enter the time employee worked in hr 45

Enter the time employee worked in hr 42

Enter the time employee worked in hr 41

Enter the time employee worked in hr 50

Enter the time employee worked in hr 51

Enter the time employee worked in hr 52

Enter the time employee worked in hr 53

Enter the time employee worked in hr 54

Enter the time employee worked in hr 55

Total Overtime Pay Of 10 Employees Is 1020.

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Select the properties and typical applications for the high carbon steels.
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Answer:

<u>Option-(A)</u>

Explanation:

<u>Typical applications for the high carbon steels includes the following;</u>

It is heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.

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Manufacturers frequently make choices about their suppliers of raw materials based on their impact on society and the environmen
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An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Giv
Anika [276]

The question is incomplete. The complete question is --

An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Given the rod has a diameter of 1.01 cm and the young's modulus of aluminum is 69 GPa, determine the following :

1. the interference value, i

2. the radial pressure at the interference of the collar and the rod

3. the maximum effective stress in the collar

4. if the yield strength of aluminium is 200 MPa and assume a safety factor of 1.5, will the aluminium collar break

Solution:

Given :

Inner radius of the collar = 1 cm

So, inner diameter, $d_1$ = 2 cm

Outer radius of the collar = 12 cm

So, outer diameter, $d_2$ = 4 cm

The aluminium rod diameter, d = 1.01 cm

Now, from the figure, we can see that there will be no interference and so the rod will easily insert inside the collar.

1. So, the interference , i =0

2. The radial pressure is also 0.

3. There will be no stress developed. So the maximum effective stress is 0

4. The collar will not break

8 0
3 years ago
Air flows at 45m/s through a right angle pipe bend with a constant diameter of 2cm. What is the overall force required to keep t
HACTEHA [7]

Answer:

b)1.08 N

Explanation:

Given that

velocity of air V= 45 m/s

Diameter of pipe = 2 cm

Force exerted by fluid  F

F=\rho AV^2

So force exerted in x-direction

F_x=\rho AV^2

F_x=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2

F=0.763 N

So force exerted in y-direction

F_y=\rho AV^2

F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2

F=0.763 N

So the resultant force R

R=\sqrt{F_x^2+F_y^2}

R=\sqrt{0.763^2+0.763^2}

R=1.079

So the force required to hold the pipe is 1.08 N.

3 0
4 years ago
Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 21 m3/min and exits at 12 b
omeli [17]

Answer:

- 46.5171kW

Explanation:

FIrst, the value given:

P1 = 1.05 bar (Initial pressure)

P2 = 12 bar (final pressure)

Heat transfer, Q = - 3.5 kW (It is negative because the compressor losses heat to the surroundings)

Mgaseous nitrogen = Mair = 28.0134 Kg/mol (constant)

Universal gas constant, Ru = 8.3143 Kj/Kgmolk

Specific gas constant, R = 0.28699 Kj/KgK

Initial temperature, T1 = 300 K

Final temperature, T2 = 400 K

Finding the volume:

P1V1 = RT1

V1 = RT1 ÷ P1

= (0.28699 Kj/KgK X 300k) ÷ 105

Note convert bar to Kj/Nm by multiply it by 100

V1 =  0.81997 m3/Kg

To get the mass flow rate:

m = volumetric flow rate / V1

= (21 m3/min x 1/60seconds) ÷ 0.81997 m3/Kg

= 0.4268Kg/s

Using tables for the enthalpy,

hT1 = 300.19 KJ/Kg

hT2 = 400.98 KJ/Kg

The enthalpy change = hT2 - hT1

= 100.79 KJ/Kg

Power, P = Q - (m X enthalpy change)

= - 3.5 - (0.4268 X 100.79)

= - 46.5171kW

3 0
3 years ago
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