Lighting a match
hope the answer helped!
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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As one distance is including twice, we can use the formula,
S = 2s1s2 / s1+s2
S = 2*3*5 / 3+5
S = 30 / 8
S= 3.75 m/s
So, your final answer is 3.75 m/s
Hope this helps!
Answer:
i dont know but what i could say is you could go on and ask the question and it would help you
Explanation:
Answer:
Explanation:
We have the relation of continuity equation as for compressible fluid flow:
where:
density of the fluid
A = area of cross-section of the fluid flow
v = velocity of flow
For a compressible fluid, the density varies.
It is also called mass flow rate, having unit as kilogram per second and is proportional to density.
The higher the density more is the mass flow rate of the fluid.