You need to explain it more simple as everyone is clueless
Hi, you haven't provided the programing language in which you need the code, I'll just explain how to do it using Python, and you can apply a similar method for any programming language.
Answer:
1. def pyramid_volume(base_length, base_width, pyramid_height):
2. volume = base_length*base_width*pyramid_height/3
3. return(volume)
Explanation step by step:
- In the first line of code, we define the function pyramid_volume and it's input parameters
- In the second line, we perform operations with the input values to get the volume of the pyramid with a rectangular base, the formula is V = l*w*h/3
- In the last line of code, we return the volume
In the image below you can see the result of calling the function with input 4.5, 2.1, 3.0.
Answer:
Explanation:
We use kinetic friction when a body is moving i.e.
for calculations.
Static friction is used when a body is in rest while kinetic friction is used when a body is moving and its value is quite low as compared to static friction .
Static friction value increases as we apply more force while kinetic friction occurs when there is relative motion between bodies.
Answer:
It will be equivalent to 338.95 N-m
Explanation:
We have to convert 250 lb-ft to N-m
We know that 1 lb = 4.45 N
So foe converting from lb to N we have to multiply with 4.45
So 250 lb = 250×4.45 =125 N
And we know that 1 feet = 0.3048 meter
Now we have to convert 250 lb-ft to N-m
So 
So 250 lb-ft = 338.95 N-m
Answer:
a). Work transfer = 527.2 kJ
b). Heat Transfer = 197.7 kJ
Explanation:
Given:
= 5 Mpa
= 1623°C
= 1896 K
= 0.05 
Also given 
Therefore,
= 1 
R = 0.27 kJ / kg-K
= 0.8 kJ / kg-K
Also given : 
Therefore,
= 

= 0.1182 MPa
a). Work transfer, δW = 
![\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}](https://tex.z-dn.net/?f=%5Cleft%20%5B%5Cfrac%7B5%5Ctimes%200.05-0.1182%5Ctimes%201%7D%7B1.25-1%7D%20%20%5Cright%20%5D%5Ctimes%2010%5E%7B6%7D)
= 527200 J
= 527.200 kJ
b). From 1st law of thermodynamics,
Heat transfer, δQ = ΔU+δW
= 
=![\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W](https://tex.z-dn.net/?f=%5Cleft%20%5B%20%5Cfrac%7B%5Cgamma%20-n%7D%7B%5Cgamma%20-1%7D%20%5Cright%20%5D%5Ctimes%20%5Cdelta%20W)
=![\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200](https://tex.z-dn.net/?f=%5Cleft%20%5B%20%5Cfrac%7B1.4%20-1.25%7D%7B1.4%20-1%7D%20%5Cright%20%5D%5Ctimes%20527.200)
= 197.7 kJ