Answer:
trail mix, soup, and gold
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Explanation:
(a) 5.66 m/s
The flow rate of the water in the pipe is given by
where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
So we can re-arrange the equation to find the speed of the water:
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
where we have
and where 
is the cross-sectional area of the pipe at the second point.
Solving for A2,
And finally we can find the radius of the pipe at that point:
Use the distance swan and the time elapsed in that interval.
Average velocity = distance / time
Average velocity = [4.0 m + 3.0m] / 3.2 s = 2.1875 m/s
Answer:
796.18 Hz
Explanation:
Applying,
Maximum velocity = Amplitude×Angular velocity
Therefore,
V' = A(2πf)............... Equation 1
Where V' = maximum velocity of the eardrum, A = Amplitude of vibration of the eardrum, f = frequency of the eardrum vibration, π = pie
make f the subject of the equation
f = V'/2πA................ Equation 2
From the question,
Given: V' = 3.6×10⁻³ m/s, A' = 7.2×10⁻⁷ m,
Constant: 3.14.
Substitute these values into equation 2
f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)
f = 796.18 Hz
