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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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Answer:

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Using trigonometry

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Differentiating with respect to t we get

\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft

Now y = 2500 ft

cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974

\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s

The angle is changing at 0.0031792338 rad/s

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3 years ago
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yarga [219]
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icebergs float on water because ice is less dense than water. The same is true for a boat: a boat floats on water because, overall, it is less dense than the water.

Explanation:

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