Answer:
For now the answer to this question is only for partial fraction. Find attached.
Answer:
hello your question is incomplete attached below is the complete question
A) overall mean = 5.535, standard deviation ≈ 0.3239
B ) upper limit = 5.85, lower limit = 5.0
C) Not all the samples meet the contract specifications
D) fluctuation ( unstable Asphalt content )
Explanation:
B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday
The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85
The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0
attached below is the required plot
C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :
15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20
D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed
Answer:
a) Two points that differ in phase by π/8 rad are 0.0461 m apart.
b) The phase difference between two displacements at a certain point at times 1.6 ms apart is 4π/3.
Explanation:
f = 420 Hz, v = 310 m/s, λ = wavelength = ?
v = fλ
λ = v/f = 310/420 = 0.738 m
T = periodic time of the wave = 1/420 = 0.00238 s = 0.0024 s = 2.4 ms
a) Two points that differ in phase by π/8 rad
In terms of the wavelength of the wave, this is equivalent to [(π/8)/2π] fraction of a wavelength,
[(π/8)/2π] = 1/16 of a wavelength = (1/16) × 0.738 = 0.0461 m
b) two displacements at times 1.6 ms apart.
In terms of periodic time, 1.6ms is (1.6/2.4) fraction of the periodic time.
1.6/2.4 = 2/3.
This means those two points are 2/3 fraction of a periodic time away from each other.
1 complete wave = 2π rad
Points 2/3 fraction of a wave from each other will have a phase difference of 2/3 × 2π = 4π/3.
Answer:
2.65 MPa
Explanation:
To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

<u>Where:</u>
p: is the gage pressure = 108 kPa
r: is the inner radius of the ball
t: is the thickness = 3 mm
Hence, we need to find r, as follows:

<u>Where:</u>
d: is the outer diameter = 300 mm

Now, we can find the normal stress (σ) in the wall of the basketball:
Therefore, the normal stress is 2.65 MPa.
I hope it helps you!