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Question Number 55288 by peter frank last updated on 20/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Feb/19

$$F=\left[{MLT}^{-2}\right]A=\left[L^2\right]\rho =\left[{ML}^{-3}\right]V=\left[{LT}^{-1}\right]$$$$$$$$k=numificalconstant$$$$F=kA\rho V^{\alpha }$$$${MLT}^{-2}=k\times L^2\times {ML}^{-3}\times L^{\alpha }{T}^{-\alpha }$$$${MLT}^{-2}={kML}^{2+\alpha -3}{T}^{-\alpha }$$$$\alpha -1=1\alpha =2\to \left(comparingpowerofL\right)$$$$T^{-2}=T^{-\alpha }so\alpha =2$$$$so\alpha =2$$$$$$$$$$

Commented by peter frank last updated on 20/Feb/19

$$\mathrm{thank}\mathrm{you}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Feb/19

$$F=kA\rho V^2$$$$mg=kA\rho V^2$$$$k=\frac{mg}{A\rho V^2}=\frac{1\times 10}{4\pi {\left(50\times {10}^{-3}\right)}^2\times {\rho }_{air}\times {\left(111\right)}^2}$$$$plscalculatebyyourself...$$

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