Question #e2ec5

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Question #e2ec5

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Answer 1 · 2017-05-27T05:41:40

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Explanation:

P = force/area $= \frac{k g m s^{- 2}}{m^{2}} = k g m^{- 1} s^{- 2}$ $=(kgms^-2)/m^2= kgm^-1s^-2$
v $= m s^{- 1}$ $= ms^-1$
A $= m^{2}$ $=m^2$

$Q = \frac{k g m^{- 1} s^{- 2}}{m s^{- 1} m^{2}}$ $Q= (kgm^-1s^-2)/(ms^-1 m^2)$

$Q = k g m^{- 1 - 3} s^{- 2 - (- 1)}$ $Q=kgm^(-1-3)s^(-2-(-1)$

$Q = k g m^{- 4} s^{- 1}$ $Q=kgm^-4s^-1$

Answer 2 · 2017-05-27T05:42:12

$K g m^{- 4} {sec}^{- 1}$ $Kgm^-4sec^-1$

Explanation:

See , ATQ

$Q = \frac{P}{v \cdot A}$ $Q=(P)/(v*A)$
$\Rightarrow \frac{\frac{F}{A}}{\frac{m}{s} \cdot A}$ $=>(F/A)/(m/s*A)$
$\Rightarrow \frac{F}{v \cdot A^{2}}$ $=>(F)/(v*A^2)$
$\Rightarrow \frac{M L T^{- 2}}{L T^{- 1} \cdot L^{5}}$ $=>(MLT^(-2))/(LT^(-1)*L^5)$

$\Rightarrow \frac{M L T^{- 2}}{L^{5} T^{- 1}} = M L^{- 4} T^{- 1}$ $=>(MLT^-2)/(L^5T^-1)=ML^-4T^-1$

hence , equals to $K g m^{- 4} {sec}^{- 1}$ $Kgm^-4sec^-1$

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Question #e2ec5

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