Answer:
A) 14.75
B) 3.36Kj
C) 2384.2k
D) 117.6kW
E) 57.69%
Explanation:
Attached is the full solutions.
Answer:
c. an abrupt increase followed by a gradual decrease
Explanation:
At the headwater, the flow gradient starts high but then slowly decreases as the river moves downstream to its mouth.
Answer:
86701 Micrometers.
Explanation:
Multiply 0.86701 dm by 100,000 to get 86701 um.
Answer:
Velocity of ball B after impact is
and ball A is 
Explanation:
= Initial velocity of ball A

= Initial velocity of ball B = 0
= Final velocity of ball A
= Final velocity of ball B
= Coefficient of restitution = 0.8
From the conservation of momentum along the normal we have

Coefficient of restitution is given by



Adding the above two equations we get



From the conservation of momentum along the plane of contact we have


Velocity of ball B after impact is
and ball A is
.
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