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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A. Her mass is 65kg
b. Her weight on Earth is 650N
c. She would weigh 240.5N on Mars
While gravitational pull pulls us down the opposing force pusshes us upward and cancels each other out
Answer:
1) Determine the domain of the following functions: d ... 3) If g(x) = x + 3 and f(x)= x² – 2x, find the value of f(g(a)). ... 6) Given the graph of f(x) to the right, determine: ... 8) Given f(x)= x? and g(x)= 2* The inverse of g is a function, but the inverse off is ... -3(x-1)= -5 4 (-3). -3x+ 3 = y. 10) The graph of a function f (x) is given at the.
Explanation:
Correct me if I’m wrong but I think it’s C