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If the mass of a system is halved (m’=m/2) and the net force acting on it is doubled (F’=2F), then its acceleration would be:* 7 points unchanged (x1) halved (x0.5) quartered (x0.25) doubled (x2) quadrupled (x4)

If the mass of a system is halved mm2 and the net force acting on it is doubled F2F then its acceleration would be 7 points unchanged x1 halved x05 quartered x0 class=

Respuesta :

Given:

The mass of the system is m' =m/2

The net force acting on it is F' = 2F

To find the acceleration.

Explanation:

The acceleration will be

[tex]\begin{gathered} a^{\prime}=\frac{F^{\prime}}{m^{\prime}} \\ =\frac{2F}{\frac{m}{2}} \\ =\frac{4F}{m} \\ a^{\prime}=4a \end{gathered}[/tex]

Thus, the acceleration would be quadrupled(x4).

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