Answer:
they are important together, but if you want to use just one future you must think about which one is first needed. and then try to learn for economical so don't use more money
9. ¿Cómo ha cambiado su español? ¿Cómo reacciona a estos cambios?
Answer:
diameter= 1 in
Explanation:
Some authors have developed studies to find the recommended flow rate at the pump outlet, in order to avoid high pressure losses and cavitation.
Assuming that this pump carries water, the recommended speed is 2m / s, so knowing the flow rate = 20gpm we can find the diameter of the pipe.
Q=flow rate=20gpm=0.0012618m^3/s
V=speed=2m/s
A=Area
Q=VA
A=Q/V
![A=\frac{0.0012618m^3/s}{2m/s} =0.0006309m^2=630.2mm^2\\\\](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B0.0012618m%5E3%2Fs%7D%7B2m%2Fs%7D%20%3D0.0006309m%5E2%3D630.2mm%5E2%5C%5C%5C%5C)
Taking into account that the flow is through a circular pipe we can use the equation to find the area of a circle, and then find the diameter
![A=\frac{\pi D^2 }{4} \\D=\sqrt[2]{\frac{4A}{\pi } } =\sqrt{\frac{4(630.8)}{\pi } } =23.34mm](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20D%5E2%20%7D%7B4%7D%20%5C%5CD%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B4A%7D%7B%5Cpi%20%7D%20%7D%20%3D%5Csqrt%7B%5Cfrac%7B4%28630.8%29%7D%7B%5Cpi%20%7D%20%7D%20%3D23.34mm)
the result is 23.34mm, if we find the closest commercial diameter to this value, we find that it is 25.4mm = 1in
Answer:
Explanation:
def is_leap_year(year):
if(year % 400 == 0):
return True
elif year % 100 == 0:
return False
elif year%4 == 0:
return True
else:
return False
if __name__ == '__main__':
n = int(input())
if(is_leap_year(n)):
print(n,"- leap year")
else:
print(n, "- not a leap year")
check the attachment for the output