To solve this given problem, we make use of the formula:
A = Ao e^( − b t / 2 m)
Substituting all the given values into the equation:
A / Ao = e^( − b t / 2 m)
When A / Ao = 0.60 and t = 50 s, we find for b:
0.60 = e^( − b t / 2 m)
ln ( 0.60 ) = − b t / 2 m
b = − ( 2 m ) ln ( 0.60 ) / t
b = ( − 2 ) ( .200 ) ln ( 0.60 ) / 50
b = .00409
When A / Ao = 0.30, we find for t:
0.30 = e^( − b t / 2 m)
ln ( 0.30 ) = − (0.00409) t / 2 (0.200)
t = − (0.200) ln ( 0.30 ) / 0.00409
t = 118 s
Therefore the number of oscillations is:
oscillations = f * t = 2 s^-1 (118 s) = 236
Answer: 236 oscillations
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Answer:
If I remembered correctly the answer would be Ne and Ar(Neon and Argon
Answer:
When t = 0, the train has a speed of 8 m/s, which is increasing at 0.5 m/s^2. Determine the magnitude of the acceleration of the engine when it reaches point A, at t = 20 s. Here the radius of curvature of the treks is 400 m
= 0.9519 m/s²
≅ 0.95 m/s²
Explanation:
see the attached file
Hint: The North American Plate and Eurasian Plate are moving away from each other.
