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Wewaii [24]
3 years ago
15

At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. Th

e spring with a force constant 37.0 N/cm and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 66.0 kg are pushed against the other end, compressing the spring 0.370 m. The sled is then released with zero initial velocity.
What is the sled's speed when the spring is still compressed 0.180 m?

Physics
2 answers:
jekas [21]3 years ago
8 0

Answer:

2.2 m/s

Explanation:

<u>solution:</u>

To calculate change in stored energy at desired extension

ΔU = 1/2*k*(δx)^2

     = 1/2*3700*(0.37^2-0.180^2)

     = 201 N.m

use work energy theorem

ΔU = ΔK = 1/2*m*v^2 = 201

              = 2.2 m/s

<u>note:</u>

calculation maybe wrong but method is correct.

sweet [91]3 years ago
5 0

Explanation:

Below is an attachment containing the solution.

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F=ma
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8 0
3 years ago
Suppose that a steel bridge, 1000 m long, were built without any expansion joints. Suppose that only one end of the bridge was h
Stels [109]

Answer:

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Explanation:

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Put the value into the formula

\Delta L=1000\times10.5\times10^{-6}(40^{\circ}-0^{\circ})

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Explain the relationship between the current output of the power supply and the current through each component in the parallel c
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2 years ago
Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?
MrMuchimi
 T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r 

where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance. 
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Point 1: 
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