Answer:
1.74 km
Explanation:
We are given that
Speed of airplane,v=320 m/s
We have to find the smallest radius allowable for the vertical circle if the pilot's apparent weight is not exceed 7.0 times his true weight.
Let true weight =w
Apparent weight=7w
According to question
[tex]7w=w+\frac{mv^2}{r}[/tex]
[tex]7w-w=\frac{mv^2}{r}[/tex]
We know that
[tex]m=\frac{w}{g}[/tex]
[tex]6w=\frac{wv^2}{rg}[/tex]
[tex]6=\frac{v^2}{rg}[/tex]
[tex]r=\frac{v^2}{6g}[/tex]
Where [tex]g=9.8 m/s^2[/tex]
[tex]r=\frac{(320)^2}{6\times 9.8}= 1741.5 m[/tex]
[tex]r=\frac{1741.5}{1000}=1.74 km[/tex]
Where 1 km=1000 m
