Asked by ali
a hockey puck rebounds from a board. the puck is in contact with the board for 2.5 ms. dertermine the acceleration of the puck over the time interval.
V(i) = 26 m/s...at angle of 22 degrees.
V(f) = 21 m/s...at angle of 22 degrees.
a=?
V(i) = 26 m/s...at angle of 22 degrees.
V(f) = 21 m/s...at angle of 22 degrees.
a=?
Answers
Answered by
drwls
Is the 22 degrees measured from the normal to the board?
Whatever the answer to that question is, determine the change in the component of velocity normal to the board
It will be either sin 21*47 m/s or cos21*47 m/s.
Divide that by the contact tiome for the acceleration normal to the board.
There will also be a deceleration along the direction of the board (since the velocity drops), but it will be much smaller.
Whatever the answer to that question is, determine the change in the component of velocity normal to the board
It will be either sin 21*47 m/s or cos21*47 m/s.
Divide that by the contact tiome for the acceleration normal to the board.
There will also be a deceleration along the direction of the board (since the velocity drops), but it will be much smaller.
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