Question

Water flows horizontally through a garden hose with an inner diameter of .012 m at a speed of 7.8 m/s. it exits out a small nozzle with a diameter of only 0.0085 m. how fast it is travelling out of the nozzle?

Answer 1

Explanation:

The given data is as follows.

Speed of water , $v_{1}$ = 7.8 m/s

 Inner diameter , $d_{1}$ = 0.012 m

 Radius, $r_{1}$ = 0.006 m

 Inner diameter for the second nozzle, $d_{2}$ = 0.0085 m

Radius, $r_{2}$ = 0.00425

Now, let us assume that speed of water through the second nozzle be $v_{2}$.

Hence, using the equation of continuity  we will find the speed of water through the nozzle as follows.

          $Area1 \times v_{1} = Area 2 \times v_{2}$

   $\pi \times {(0.006)}^2 \times 7.8 m/s = \pi \times {(0.00425)}^{2} \times v_{2}$

             0.0002808 = $0.00001806 v_{2}$

                $v_{2}$ = 15.54 m/s

Thus, we can conclude that water is travelling at a speed of 15.54 m/s.

Answer 2

Answer: 110.12 m/s

We will use the formula A1V1 = A2V2 where 7.8 m/s is divided with 0.0085 m then multiply to 0.12 m, the result will be 110.117 or 110.12 m/s. This is related to the continuity of fluid flow in which as liquid moves horizontally, the same amount of liquid goes out as it comes in or the liquid itself do not change as it moves but the speed does when the diameter changes.