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djyliett [7]
2 years ago
7

A mass is oscillating horizontally on a spring. At the locations A, B, C, D, and E, photogates are used to measure the speed of

the mass as it moves by. The speed is used to calculate the kinetic energy ( K E ) and the location of the photogates is used to calculate the elastic potential energy ( P E elastic ) . Interpret the chart to determine the missing K E and P E elastic values. Assume all values are exact.

Physics
1 answer:
svetoff [14.1K]2 years ago
4 0

Complete Question

The complete question is is shown on the first uploaded

Answer:

The elastic potential energy at point B is  PE_{elastic} = 50J

The kinetic energy at point D is KE = 75J

Explanation:

Looking at the given point we can observe that mechanically energy(i.e potential and kinetic energy ) is conserved and it value is E_ {m} = 100J

     So at point B

           E_{m} = PE_{elastic} +KE

           100 = PE_{elastic} + KE

   KE at point B is  50J

So     PE_{elastic} = 100 - 50 = 50J

     Now at point D

          E_{m} = PE_{elastic} +KE

    PE_{elastic} at point D is 25J

 So  KE = 100 - 25 = 75J

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and T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K

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T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K

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"For a first order instrument with a sensitivity of .4 mV/K and a time" constant of 25 ms, find the instrument’s response as a f
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Answer:

Explanation:

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Plot the response y(t) as a function of time.

The plot of y(t) as a function of time can be seen in the diagram  attached below.

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170.28 mV - 109.2 mV = 80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

61.08 mV =  80 ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

0.7635  mV = ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

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t = 0.036 s

t = 36 ms

The error fraction = \dfrac{189.2-170.28  }{189.2}

The error fraction = 0.1

The error fraction = 10%

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