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Arent the area supposed to be in m^2?Aren't you suppose to multiply Hcos(8) with something with meter as unit to make it m^2 ?So that when you solve for q you'll get m^3/hr as unit?
Exactly what I was thinking
We assume the width of the cross section is 1 meter and multiply it with Hcos(8) to obtain the area in m^2
@@mauricemuleba5179 thank you very much
Well this was very helpful 🎉
what book is this?
Aoa its PRINCIPLES OF GEOTECHNICAL ENGINEERING Ninth Edition
❤
Arent the area supposed to be in m^2?
Aren't you suppose to multiply Hcos(8) with something with meter as unit to make it m^2 ?
So that when you solve for q you'll get m^3/hr as unit?
Exactly what I was thinking
We assume the width of the cross section is 1 meter and multiply it with Hcos(8) to obtain the area in m^2
@@mauricemuleba5179 thank you very much
Well this was very helpful 🎉
what book is this?
Aoa its PRINCIPLES OF GEOTECHNICAL ENGINEERING Ninth Edition
❤