Answer:
3.76
Explanation:
We are given that
Terminal speed in the spread -eagle position,[tex]v_t=165 km/h[/tex]
Terminal speed in the nosedive position,[tex]v'_t=320km/h[/tex]
We have to find the ratio of the effective cross-sectional area A in the slower position to that in the faster position.
We know that
Area, A=[tex]\frac{2mg}{C\rho v^2_t}[/tex]
[tex]A_{slower}=\frac{2mg}{C\rho(165)^2}[/tex]
[tex]A_{faster}=\frac{2mg}{C\rho(320)^2}[/tex]
[tex]\frac{A_{slower}}{A_{faster}}=\frac{(320)^2}{(165)^2}[/tex]
[tex]\frac{A_{slower}}{A_{faster}}=3.76[/tex]
Answer:
3.76.
Explanation:
terminal velocity in spread eagle position = 165 km/h
terminal velocity in nose dive position = 320 km/h
ratio of cross section area = ?
terminal velocity of the diver is given by
[tex] v = \sqrt{\dfrac{2mg}{C_{\rho}A}}[/tex]
A is the area of cross section
On rearranging
[tex] A = \dfrac{2mg}{C_{\rho}v^2}[/tex]
From the above expression we can say that cross sectional area is inversely proportional to velocity.
[tex]\dfrac{A_{slower}}{A_{faster}}=\dfrac{v^2_{faster}}{v^2_{slower}}[/tex]
[tex]\dfrac{A_{slower}}{A_{faster}}=\dfrac{320^2}{165^2}[/tex]
[tex]\dfrac{A_{slower}}{A_{faster}}=3.76[/tex]
Hence, the area ratio is 3.76.
