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Author Question: A 12.0-N object is oscillating in simple harmonic motion at the end of an ideal vertical spring. ... (Read 141 times)

LCritchfi

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on: Jul 29, 2018
A 12.0-N object is oscillating in simple harmonic motion at the end of an ideal vertical spring.
 
  Its vertical position y as a function of time t is given by y(t) = 4.50 cm cos(19.5 s-1)t - /8.
  (a) What is the spring constant of the spring?
  (b) What is the maximum acceleration of the object?
  (c) What is the maximum speed that the object reaches?
  (d) How long does it take the object to go from its highest point to its lowest point?

Question 2

The decay rate of a radioactive sample decreases from 175 kBq to 12.4 kBq in 18.3 s. What is the half-life of this sample?
 
  What will be an ideal response?


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b614102004

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Reply #1 on: Jul 29, 2018
Answer to Question 1

(a) 466 N/m (b) 17.1 m/s2 (c) 0.878 m/s (d) 0.161 s

Answer to Question 2

4.79 s


LCritchfi

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Reply 2 on: Jul 29, 2018
Wow, this really help

flexer1n1

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Reply 3 on: Yesterday
Great answer, keep it coming :)

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