Physics question? I have the answer just don't understand it?

volume flow in = volume flow out

volume flow = area*velocity

Area0*v0=Area1*v1

Areo0/Area1*v0=v1

0.75^2/0.05^2*3ft/s = 675ft/s

Oh, for the conceptual idea, what I'm using is conservation of mass. For a liquid, the density is going to be about the same in most situations. The amount of mass going in MUST equal the amount of mass exiting, and since the density is constant, the volume in is the same as the volume out. Just convert all that to rates and you'll get what I put.

I think you mean "at what speed does it leave the nozzle", otherwise the question is far too easy!

The overall flow (volume per second) is the same through the hose and through the nozzle.

Therefore, the speed of flow through the nozzle is faster than the speed of flow through the hose by a ratio equal to the ratio of the cross-sectional area of the hose and the cross-sectional area of the nozzle.

Since cross-sectional area of a circle increases as the square of the diameter, tha means:

Fn = Fh * Dh^2 / Dn^2 = Fh * (Dh / Dn)^2

Fn = flow rate through nozzle

Fh = flow rate through hose

Dn = nozzle diameter

Dh = hose diameter

The actual answer is 3 * 225 = 675 ft/s

The answer to this question is based on the ratio of the cross-sectional area the water has available to flow through.

If you pick a point in the hose, 3 feet worth of water flows past that point in one second. Since water does not compress, this means that the amount of water in 3 feet of the hose (about a cup and a half) has to flow through that VERY TINY nozzle in one second also. Since the diameter (and area) of the nozzle is much smaller, the water must flow through the nozzle much faster to get that same amount of water past it in one second that goes past a point in the hose. in that same second. This can be determined by a simple ratio, or

Area of hose x Speed of water through hose = Area of nozzle x Speed of water through nozzle.

In the main hose, the diameter is .75 inches. That is a pretty normal size for a garden hose. The area is:

A = pi x r^2

A = 3.14 x (d/2)^2

A = 3.14 x (.375)^2 or,

A = 3.1416 x .14

A = .442 square inches for the hose.

For the nozzle we use the exact same formula with the diameter of the nozzle being .050 in, which is a VERY small diameter for a nozzle, about the size of the "o" in nozzle on your computer screen. We get;

A (nozzle) = .00196 sq in.

Using the ratio we determined above, we get

A (hose) x S1 (speed of water) = A (nozzle) x S2 (speed of water) Substitute in the numbers we found to get;

.442 x 3 = .00196 x S2 or

1.32 = .00196 x S2

1.32/.00196 = S2

676ft/sec = S2

Since we are cramming about a cup and a half of water through a hole the size of this "o" in ONE SECOND, it makes sense that it would move pretty fast. Still, I would call that unreasonably fast for a garden hose, and I suspect the pressure necessary to do this would destroy the hose if we actually tried it.

I hope this helped!

Ratio between the area of the hose to area of the nozzle

= 0.140625 : 0.000625 (The square of the radii)

Now just multiply the speed of flow by that ratio...

3.0 X 0.140625 / 0.000625

= 675ft/s (or 680 to 2 sig figs)