Answer:
Final length= 746.175 mm
Explanation:
Given that Length of aluminium at 223 C is 750 mm.As we know that when temperature of material increases or decreases then dimensions of material also increases or decreases respectively with temperature.
Here temperature of aluminium decreases so the final length of aluminium decreases .
As we know that

Now by putting the values

ΔL=3.82 mm
So final length =750-3.82 mm
Final length= 746.175 mm
Answer:
b) Commonly used to transmit network signals over great distances.
Explanation:
The transmission of information or data by using microwave radio waves is known as microwave transmission. Microwave transmitter is commonly used to transmit network signals over great distances. It is an electronic device that transmits and receives radio frequency signals ranging from 1GHz to 100GHz.
The microwave transmitter has a wide range of applications and these includes, radio stations, television stations, mobile phones, radio astronomy, radar,
Answer:
If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
Explanation:
Reynolds number is an important dimensionless parameter in fluid mechanics.
It is calculated as;

where;
ρ is density
v is velocity
d is diameter
μ is viscosity
All these parameters are important in calculating Reynolds number and understanding of fluid flow over an object.
In aerodynamics, the higher the Reynolds number, the lesser the viscosity plays a role in the flow around the airfoil. As Reynolds number increases, the boundary layer gets thinner, which results in a lower drag. Or simply put, if Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
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:
Following is attached the solution or the question given.
I hope it will help you a lot!
Explanation: