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LUCKY_DIMON [66]
2 years ago
8

A karate master strikes a board with an initial velocity of 10.0 m/s, decreasing to 1.0 m/s as his hand passes through the board

. If the time of contact with the board is 0.002 0 s, and the mass of the coordinated hand and arm is 1.0 kg, what is the force exerted on the board
Physics
1 answer:
kenny6666 [7]2 years ago
6 0

The force exerted on the board by the karate master given the data is -4500 N

<h3>Data obtained from the question </h3>
  • Initial velocity (u) = 10 m/s
  • Final velocity (v) = 1 m/s
  • Time (t) = 0.002 s
  • Mass (m) = 1 Kg
  • Force (F) = ?
<h3>How to determine the force</h3>

The force exerted can be obtained as illustrated below:

F = m(v - u) / t

F = 1 (1 - 10) / 0.002

F = (1 × -9) / 0.002

F = -4500 N

Learn more about momentum:

brainly.com/question/250648

#SPJ1

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