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3 years ago

What is the total kinetic energy of a 2500 lbm car when it is moving at 80 mph (in BTU)?

Engineering

1 answer:

Damm [24]3 years ago

3 0

Answer:

The kinetic energy will be 687.186 BTU

Explanation:

We have given mass of car = 2500 lbm

We know that 1 lbm = 0.4535 kg

So 2500 lbm = $2500\times 0.4535=1133.75kg$

Speed = 80 mph

We know that 1 mile = 1609.34 meter

1 hour = 3600 sec

So $80mph=\frac{80\times 1609.34}{3600}=35.763m/sec$

We know that kinetic energy $E=\frac{1}{2}mv^2=\frac{1}{2}\times 1133.75\times 35.76^2=725.033KJ$

We know that 1 KJ = 0.9478 BTU

So 725.033 KJ = 725.033×0.9478 = 687.186 BTU

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3 years ago

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

fgiga [73]

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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2 years ago

An aluminium alloy tube has a length of 750 mm at a temperature of 223°C. What will be its length at 23°C if its coefficient of

uranmaximum [27]

Answer:

Final length= 746.175 mm

Explanation:

Given that Length of aluminium at 223 C is 750 mm.As we know that when temperature of material increases or decreases then dimensions of material also increases or decreases respectively with temperature.

Here temperature of aluminium decreases so the final length of aluminium decreases .

As we know that

$\Delta L=L\alpha\Delta T$

Now by putting the values

$\Delta L=750\times \25.5\times 10^{-6}\times 200$

ΔL=3.82 mm

So final length =750-3.82 mm

Final length= 746.175 mm

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