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baherus [9]
2 years ago
8

could anyone determinate the period knowing that it performs 4000 vibrations in 0.5 minutes, Sorry my english is bad

Physics
1 answer:
dexar [7]2 years ago
4 0

Answer:

f = 4000 / 30 sec = 133.3    vibrations/sec

P = 1 / f = .0075 sec       period of 1 vibration

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4 years ago
I need to know the correct answer please ?
mrs_skeptik [129]

Answer:

675 Pa.

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F = 5+2cos(15t) kN

Area (a) = 8*10-3 m2

Now at t =4 sec

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4 years ago
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mr_godi [17]

The answer is; C


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