Answer:
84%
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
[tex]y_0=1\ m[/tex]
x = 10 m
t = Time taken
[tex]v_0[/tex] = 3.5 m/s (assumed, as it is not given)
[tex]A_0=\pi 1.5^2[/tex]
We have the equation
[tex]y=y_0+ut+\dfrac{1}{2}gt^2\\\Rightarrow 0=y_0-\dfrac{1}{2}gt^2\\\Rightarrow t=\sqrt{\dfrac{2y_0}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1}{9.81}}\\\Rightarrow t=0.45152\ s[/tex]
[tex]x=x_0+vt\\\Rightarrow 10=0+v0.45152\\\Rightarrow v=\dfrac{10}{0.45152}\\\Rightarrow v=22.14741\ m/s[/tex]
From continuity equation we have
[tex]Av=A_0v_0\\\Rightarrow A=\dfrac{A_0v_0}{v}\\\Rightarrow A=\dfrac{\pi 1.5^2\times 3.5}{22.14741}\\\Rightarrow A=1.11706\ cm^2[/tex]
Fraction is given by
[tex]f=\dfrac{A_0-A}{A_0}\times 100\\\Rightarrow f=\dfrac{\pi 1.5^2-1.11706}{\pi 1.5^2}\times 100\\\Rightarrow f=84.196\ \%\approx 84\ \%[/tex]
The fraction is 84%
A fraction of the cross-sectional area of the hose hole that Isabella must cover is 84%.
Given the following data:
Vertical distance = 1.0 meters.
Horizontal distance = 10.0 meters.
Radius = 1.5 cm.
Vertical speed = 3.5 m/s.
The time taken to reach a maximum height is given by this formula:
[tex]H=\frac{1}{2} gt^2\\\\t=\sqrt{\frac{2H}{g} } \\\\t=\sqrt{\frac{2\times 1.0}{9.8} }[/tex]
t = 0.452 seconds.
For the velocity required, we have:
[tex]x=x_o+Vt\\\\10=0+V0.452\\\\V=\frac{10}{0.452}[/tex]
V = 22.12 m/s.
From continuity equation, we have:
[tex]A=\frac{A_0V_0}{V} \\\\A=\frac{\pi r^2V_0}{V}\\\\A=\frac{\pi1.5^2 \times 3.5}{22.12} \\\\A=\frac{24.7433}{22.12}[/tex]
A = 1.118 cm^2.
Now, an expression for the fraction of the cross-sectional area is given by:
[tex]F=\frac{A_0-A}{A_0} \times 100\\\\F=\frac{7.0695-1.118}{7.0695} \times 100\\\\F=\frac{5.9515}{7.0695} \times 100\\\\F=0.8419 \times 100[/tex]
F = 84.19 ≈ 84%.
Read more on speed here: brainly.com/question/10545161
