0:00
Good day.
00:01
In this question, we will be solving the flow rate of the water per minute.
00:08
So using the bernoulli's equation, since there is a steady flow of fluid, we can use the bernoulli's equation, where bernoules equation is p1, plus one half row v1 squared, plus height 1, rog, is equal to pressure 2, plus 1, r v2 squared plus h2 ra g where the one is representing the top level and the two is the raface or the outlet of the water so since the pressure at the outlet is atmospheric we can have that pressure one is equal to pressure two so we are left with 1 half raw v1 squared plus h1 rog is equal to 1 half raw v2 squared plus h2 rog and from our height which is the height at the top level is 5 meters our height 1 is equal to 5 meters and our height 2 is 0 therefore we can cancel this group here next we are left with 1 half raw v1 squared plus h1 raw g is equal to one half row v2 squared so since the tag is large enough therefore our initial average speed would be zero we assume it to be zero since at the top there is no movement therefore we can cancel out this group here also we are left with h1 rog equals one half row v2 squared canceling the raw, we now have the value for our v2, where v2 is equal to 2g h1 minus h2.
02:52
This equation that we derive is called the torricelli's equation.
02:57
So substituting the values that we have to solve for the average speed at the arrepease, we have 2.
03:09
Times 9 .81 meters per second squared times 5 meters minus 0 is 5 meters therefore our v2 is 9 .9 meters per second...