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.

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

where 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

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it