00:01
Hi there, so for this problem, we are given that a garden hose attached with a nozzle is used to fill a 10 gallons bucket.
00:14
So we are given the volume for this that is equal to 10 gallons.
00:18
The inner diameter of the hose is also given, and that is 2 centimeters.
00:24
So let's label this as the inner radius.
00:32
Call this the radius r that is equal to 0 and 2 centimeters.
00:40
But remember, oh sorry, this is the inner diameter, okay? so the inner diameter is 2 centimeters that in meters is 0 .02 meters, okay? now, and it reduces to 0 .8 centimeters at the nozzle exits.
01:03
So we are going to call this the radius re.
01:06
But that is a diameter, okay? let's call them the e, and that is 0 .8 centimeters.
01:17
That meters is 0 .008.
01:25
Now, once we have this, and we're told that if it takes an interval of time that is equal to 50 seconds in order to fill the bucket, for part a of this problem, the question is about the volume.
01:40
And mass flow rates of water through the hose.
01:45
So, for this, we will have that the rate of change of the volume is just the volume divided by the interval of time.
01:53
So that will be the 10 gallons that we are given for this, divided by the interval of time, that is 50 seconds.
02:00
Now, we need to answer this in liters per second.
02:05
So we can multiply this by a factor of conversion.
02:07
We know that 3 .7 ,854 liters equals to one gallon.
02:16
Then from this, we obtain a value of 0 .757 in units of liters per second.
02:25
So that's the solution, one of the solutions for the first part of this problem.
02:31
Now, the mass low rate, we can calculate that through the density times the volume.
02:38
The volume rate of change...