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il63 [147K]
3 years ago
15

A spring with a force constant of 5.3 n/m has a relaxed length of 2.60 m. when a mass is attached to the end of the spring and a

llowed to come to rest, the vertical length of the shaik (rs53237) – work, energy, power – theisen – (20167) 2 spring is 3.62 m. calculate the elastic potential energy stored in the spring. answer in units of j
Physics
1 answer:
olganol [36]3 years ago
6 0
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant. 
<span> U = (1/2)kx^2 
</span><span> U = (1/2)(5.3)(3.62-2.60)^2 
</span> U = <span> <span>2.75706 </span></span>J
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What is the answer to this question?
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(15 x 1000)m/3600s/s, or (15 x 1000)m/3600 s^2

We can reduce this to:

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3 0
3 years ago
Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an
castortr0y [4]

Answer:

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  • Fy = 1.72 N
  • F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

  F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

  f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

  = -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

  ≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

  ≈ (-9.15309, 1.71982)

The resultant component forces are ...

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  • Fy = 1.72 N

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  F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

  F∠γ ≈ 9.31∠-10.6°

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Answer:

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