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Natasha2012 [34]
4 years ago
11

Lee majors, a professional stuntman, is preparing to run off a dock and dive into a large kayak. the stuntman has a mass of 80.8

kg and the kayak has a mass of 29.9kg. if lee dives horizontally with a velocity of 5.50ms... what is his velocity in ms immediately after he jumps into the kayak? to simplify things, ignore any velocities in the y-direction.
Physics
1 answer:
Len [333]4 years ago
8 0
4.01 m/s  
The is a conservation of momentum problem. The stuntman's momentum plus that of the kayak will remain constant. Since the momentum of the kayak is 0 at the beginning, we need to calculate what the momentum will be the moment after the stuntman lands in the kayak. We will consider the landing to be a non-elastic collision. 
Initial momentum 
80.8 kg * 5.50 m/s = 444.4 kg*m/s
 Total mass of stuntman and kayak
 80.8 kg + 29.9 kg = 110.7 kg
 Speed of kayak and stuntman
 444.4 kg*m/s / 110.7 kg = 4.014453478 m/s

  Rounding to 3 significant figures gives 4.01 m/s
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Answer:

X = 2146.05 m

Explanation:

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t = √2h/g   (2)

Where g is gravity acceleration which is 9.8 m/s². Replacing the data into the expression we have:

t = √(2*2500)/9.8

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Now replacing into (1) we have:

X = 95 * 22.59

<h2>X = 2146.05 m</h2>

This is the distance where the package should be released.

Hope this helps

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Which of the following is least likely to result from seafloor spreading​
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kicyunya [14]
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A sky diver, with parachute unopened, falls 625 m in 15.0 s. Then she opens her parachute and falls another 362 m in 139 s. What
Jobisdone [24]

Answer:

v_{avg} = 6.41 m/s

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Average velocity is defined as the ratio of total displacement of the motion and total time taken in that motion

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so first it took time t = 15 s to drop without open parachute

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so average velocity is given as

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v_{avg} = 6.41 m/s

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