A 0.25 Kg hockey puck is hit with a force of 7.00 N. The force of friction acting on the puck when it is hit is 0.30 N. What is the magnitude of the acceleration of the hockey puck? A 12.7 m/s2 B 28.0 m/s2 C 26.8 m/s2 D 29.2 m/s2

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A 0.25 Kg hockey puck is hit with a force of 7.00 N. The force of friction acting on the puck when it is hit is 0.30 N. What is the magnitude of the acceleration of the hockey puck? A 12.7 m/s2 B 28.0 m/s2 C 26.8 m/s2 D 29.2 m/s2

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C is the answer. just find the net force by subtracting 7-0.3 and dividing by acceleration
i dont understand what it means by magnitude of acceleration
it just means size

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what would the acceleration be in this case? would be have to solve for it?
F net= massxacceleration
\[F _{1}-F _{2}=ma\] you solve for a by dividing by the mass. F1 is 7 and F2 is .3
force - friction = mass x acceleration 7.0N - 0.3N= 0.25 x a 6.7/0.25 = a a=26.8m/s2
This is simply an instantaneous acceleration for the puck at this given moment in time.
thanks yall! sorry i wish i couldve given you all the best answer!
i just noticed the typo in my first post, its divide by mass
yeah thats why i was confused @qvbsinteta haha
F=ma magnitude of the acceleration just means that the acceleration is in the direction of the applied force and it cannot be zero (0). so just keep in mind that magnitude is the same as the size.

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