Continuity Equation problem

[jdg]

Next question: A garden hose with internal diameter of 13.5 mm lies flat on a sidewalk while water is flowing in it at a speed of 6 m/s. A person happens to step on it at the very edge of the opening of the hose and decreases its internal diameter by a factor of 9

So D (1) = 0.0135m
r (1) = 0.00675m
D (2) = 0.0135/9 = 0.0015m
r (2) = 0.00075m
A (1) = pi*r^2 = (3.14...)(0.00675^2) = 1.4134...e-4
A (2) = pi*r^2 = (3.14...)(0.00075^2) = 1.767...e-6

1. What was the water flow rate in the hose prior to the person stepping on it?
- I got this part: J(1) = A(1)V(1) = 8.59 m3/s

2. What is the flow rate of water after the person steps on it?
part 2 I did J = A(1)V(1) = A(2)V(2):

So V2 = V1*(A1/A2) = 486 m/s

Is this right? 3. What is the speed of the water just as it exits the hose after the person steps on it?

And for part 3 I did

J = (A2)(V2) = 8.59e-4 m3/s

 

[kuruman]

- I got this part: J(1) = A(1)V(1) = 8.59 m3/s

You didn't get this part. 8.59 m³/s is about 2200 gallons a second. Do you really believe that a 1/2-inch diameter hose can put out that much water?

Is this right?

It can't be right. A speed of 486 m/s is almost Mach 1.5.

It looks like you messed up your powers of 10. You also don't understand the continuity equation. When the person steps on the hose, the speed should increase, not decrease because if the area becomes smaller, the speed must increase to keep the product $Av$ constant.