Where’s the illustration?
Answer:
product differentiation
Explanation:
Based on the information provided within the question it can be said that this is an example of product differentiation. This term refers to the process that a company undergoes in order to set their product apart from the competition in order to persuade customers to choose their product. Which in this scenario Thirstoid is redisigning their label and adding sports celebrities that the competition does not have.
Answer:
Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that, through cellular respiration, can later be released to fuel the organism's activities.
Answer:
0.0002778° are equivalent to one second of arc.
Explanation:
1 degree consists of 60 arc minutes.
1 arc minutes consists of 60 arc seconds.
It can also be written as:
1 minute of arc comprise of 1/60 degree.
1 second of arc comprise of 1/60 arc minutes.
So,
<u>1 second of arc comprise of (1/60)*(1/60) degree.</u>
Thus,
Thus,
<u>0.0002778° are equivalent to one second of arc.</u>
Answer:
First compute the characteristic length and the Biot number to see if the lumped analysis is applicable
Lc = V/A = (pie*D3/6) / (pie * D2)= 1.2/6 = 0.0012/6= 0.0002m
Bi = hLc/K = (110W/m2.oC)(0.0002)(moC/35W)= 110*0.0002/35 = 0.0006 less than 0.1
Since the Biot number is less than 0.1, we can use the lumped parameter analysis.
In such an analysis, the time to reach a certain temperature is given by
t = -1/bIn(T-Tinfinite/T - Tinfinite)
From the data in the problem we can compute the parameter, b, and then compute the time for the ratio (T – Tinfinite/(Ti – Tinfinite) to reach the desired value.
b = hA/pCpV = h/pCpLc = 110/8500*0.0002 *320*s
b = 110/544s = 0.2022/s
The problem statement is interpreted to read that the measured temperature difference T – Tinfinite has eliminated 98.5% of the transient error in the initial temperature reading Ti – Tinfinite so the value of value of (T – Tinfinite)/(Ti – Tinfinite) to be used in this equation is 0.015
t = -1/bIn(T-Tinfinite/T - Tinfinite)
t = -s/0.1654 (In0.015)
t = (-s*-4.1997)/0.2022
t = 20.77s
It will take the thermocouple 20.77s to reach 98.5% of the initial temperature