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Ulleksa [173]
3 years ago
11

A squirrel runs along an overhead telephone wire that stretches from the top of one pole to the next. It is initially at positio

n xi=3.37 mxi=3.37 m , as measured from the center of the wire segment. It then undergoes a displacement of Δx=−6.83 mΔx=−6.83 m . What is the squirrel's final position xfxf ?
Physics
1 answer:
Anna007 [38]3 years ago
6 0

Answer:

- 3.46 m

Explanation:

initial position, xi = 3.37 m

displacement, Δx = - 6.83 m

Let the final position is xf.

So, displacement = final position - initial position

Δx = xf - xi

- 6.83 = xf - 3.37

xf = 3.37- 6.83

xf = - 3.46 m

Thus, the final position of the squirrel is - 3.46 m.

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1) -0.5 m/s

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Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

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2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

v^2-u^2=2as

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u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

a=-0.25 m/s^2 is the deceleration of the cannon

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The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

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Special relativity "shows up" when the speeds get very high indeed.

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