00:01
So the cross -sectional area of the hose for part a, first would be equalling pi r squared, or we can say pi multiplied by the diameter squared divided by four.
00:11
So this would be equalling pi multiplied by 2 .74 centimeters, quantity squared, divided by four.
00:24
Now, we know that the volumetric flow rate, volume per unit time, would be a times, v this would be equaling 25 .0 liters for every 1 .50 minutes so one and a half minutes and so we can say that then v is equaling 25 .0 liters divided by 1 .50 minutes and then this would be divided by a the cross -sectional area so we have that v the velocity would be 25 .0 liters divided by 1 .50 minutes.
01:08
This would be multiplied by 4 divided by pi times 2 .74 centimeters quantity squared multiplied by one minute for every 60 seconds multiplied by 10 to the third cubic centimeters for every one liter and this is giving us 47.
01:40
Centimeters per second or we can say 0 .471 meters per second.
01:49
This would be our final answer for the velocity for part a.
01:54
For part b, we can say a sub 2 divided by a sub 1.
02:01
This will be equal to essentially d sub 2 divided by d sub 1 quantity squared...