Number of times the diaphragm move back and forth is 5.59×10^4
<u>Explanation:</u>
Given data,
ω=4.6 s
we have the formula
f=ω/2π
The number of times the diaphragm moves back and forth in 4.6 s is
Number of times= ft
Number of times= ft
=(ω/2π) t
=(7.54×10^4 rad/sec)(4.6 s)/2π
Number of time=5.59×10^4
Number of times the diaphragm move back and forth is 5.59×10^4
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Answer:
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Explanation:
Answer:
Explanation:
Given
wall is 10 m from student
8 m high
Therefore student should launch at an angle \theta such that its maximum height is 8 m and its range is 20 m
For maximum height(h)
Range (R)
divide 1 & 2