Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Par
1 answer:
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Answer:
Explanation:
Two frequencies with magnitude 150 Hz and 750 Hz are given
For Pipe open at both sides
fundamental frequency is 150 Hz as it is smaller
frequency of pipe is given by
where L=length of Pipe
v=velocity of sound
for n=1
and f=750 is for n=5
thus there are three resonance frequencies for n=2,3 and 4
For Pipe closed at one end
frequency is given by
for n=0
for n=2
Thus there is one additional resonance corresponding to n=1 , between and
Answer:
See explanation below
Explanation:
As I say in the comments, the question is incomplete, however, I will try to answer this by using data that I found on another site.
This is the part of the question that is not here:
If measurements are recorded to the nearest
tenth of a kilogram, find the fraction of these poodles
with weights
(a) over 9.5 kilograms;
(b) of at most 8.6 kilograms;
So, assuming a mean of 8 kg, and 0.9 of standard deviation, let X represents the weight of the poodles
The expression to calculate the fraction of poodle needed is:
Z = X - u / d
u: weight of the large number of poodle
d: standard deviation
Replacing data of a) wer have:
Z = 9.5 - 8 / 0.9
Z = 1.67
With this value, we need to take the value of Z, and see the area under the curve of standard deviation (see table attached)
Therefore:
P (X > 9.5) = P(Z > 1.67) = 0.5 - P (Z < 1.67) = 0.5 - 0.4525 = 0.0475
b) In this part, is the same as part a) so:
Z = 8.6 - 8 / 0.9 = 0.67
The value for area in the curve is 0.2486 so:
P = 0.5 + 0.2486 = 0.7486
Hope this helps
