00:01
Hi guys in this problem we need to calculate the area of section 3 so we know that a 3 is equal pi over 4 times d3 squared d3 is the diameter of this section 3 so we know that d3 is equal to 0 .13 so this is pi over 4 times 0 .13 squared this is equal to 0 .0132 meter squared.
00:37
Okay, so with the relation to calculate, that this charge at section 3, q3, so q3 is equal to a3 times v3, v3 at the average velocity of section 3, and q to 3 is the rate flow of war at section 3.
00:56
Okay, so now we need to substitute here.
00:59
This is 0 .0132 v3 okay so then we need to calculate at the area of section 2 we have a 2 is equal to by over 4 times d2 squared so we know that d2 is 0 .18 we have buy over 4 times 0 .18 we have buy over 4 times 0 .18 squared so this is equal to 0 .02544 meter squared.
01:38
Okay, then we know that q2 is equal to a2 times v2.
01:45
So a2 is 0 .02544 times 12.
01:54
So this is equal to 0 .305 meter cube per second...