Question

Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding the hose in herhand 1.0 meters above the ground and is trying to spray Ferdinand,who is standing 10.0 meters away. I know so far that she cannotspray Ferdinand at the current position and with the curreentspeed of spray. I got stuck inthe following question:
To increase the range of the water, Isabellaplaces her thumb on the hose hole and partially covers it. Assuming that the flow remains steady, what fraction f of the cross-sectional area of the hose hole does shehave to cover to be able to spray her friend?

Assume that the cross section of the hoseopening is circular with a radius of 1.5 centimeters.

Answer 1

Answer:

Explanation:

According to the formula below, with constant flow rate, the less cross-sectional area there is, the faster water would flow, and vice-versa

$\dot{V} = A*v$

where $\dot{V} m^3/s$ is the constant flow rate,

A m2 is the cross-sectional area

v m/s is the water speed.

So if the flow rate is constant, when A decreases, v must increase proportionally.

Since this problem is missing the water speed, here are the steps to solve it

Step 1: find the new spray speed that could reach Ferdinand

Step 2: find the ratio of this new spray speed to the old one, this will also be the ratio of the old cross-sectional area to the new one.

Step 3: find the fraction f of the cross-sectional area of the hose hole