Answer:
k = 0.1118 per min
Explanation:
Assume;
Initial number of bacteria = N0
Number of bacteria IN 'T' time = Nt
So,
![Nt=N0e^{-kt}\\\\in\ 6.2 min\\\\\\frac{N0}{2}= N0e^{-k(6.2)}\\\\ln\frac{1}{2} = -k[6.2]](https://tex.z-dn.net/?f=Nt%3DN0e%5E%7B-kt%7D%5C%5C%5C%5Cin%5C%206.2%20min%5C%5C%5C%5C%5C%5Cfrac%7BN0%7D%7B2%7D%3D%20N0e%5E%7B-k%286.2%29%7D%5C%5C%5C%5Cln%5Cfrac%7B1%7D%7B2%7D%20%3D%20-k%5B6.2%5D)
k = 0.1118 per min
I believe the answer is: A. Passive heating and cooling.
Answer: They showed that many people worked on the project and demonstrate how hard they worked.
Explanation:
Answer:

Explanation:
We are given:
m = 1.06Kg

T = 22kj
Therefore we need to find coefficient performance or the cycle


= 5
For the amount of heat absorbed:

= 5 × 22 = 110KJ
For the amount of heat rejected:

= 110 + 22 = 132KJ
[tex[ q_H = \frac{Q_L}{m} [/tex];
= 
= 124.5KJ
Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c
Convert 69.5°c to K we have 342.5K
To find the minimum temperature:
;

= 285.4K
Convert to °C we have 12.4°C
From the refrigerant R -134a table at
= 12.4°c we have 442KPa