Answer:
[tex]\textrm{Dimension of C }=\ [ML^{-3}T^{0}][/tex]
Step-by-step explanation:
As given in question drag force depends upon the product of the cross sectional area of the object and the square of its velocity
and drag force can be given by
[tex]F=CAv^2[/tex] (1)
It is given that
Dimension of mass = [M]
Dimension of length = [L]
Dimension of time = [T]
So, by using above dimension we can write
the dimension of force,
[tex]F=[MLT^{-2}][/tex]
dimension of cross-section area,
[tex]A=[L^2][/tex]
and dimension of velocity
[tex]v=[LT^{-1}][/tex]
now, by putting these values in equation (1), we will get
[tex]F=CAv^2[/tex]
[tex]=>[MLT^{-2}]=C[L^2][LT^{-1}]^2[/tex]
[tex]=>C=[ML^{-3}T^0][/tex]
Hence, the dimension of constant C will be,
[tex]C=[ML^{-3}T^0][/tex]
The dimensions of C from the expression above is ML^-3
Given the function that relates drag force with the cross-sectional area of the object and the square of its velocity expressed as:
[tex]F_{air} = CAv^2[/tex]
Make C the subject of the formula to have:
[tex]C =\frac{F}{Av^2}[/tex]
Given the following dimensions
[tex]M =MLT^{-2}[/tex]
A = L²
v = [tex]LT^{-1}[/tex]
Substitute into the formula. the dimension of C will be given as:
[tex]C=\frac{MLT^{-2}}{L^2L^2T^{ -2}}\\C= ML^{-3}[/tex]
Henc the dimensions of C from the expression above is ML^-3
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