The dP/dt of the adiabatic expansion is -42/11 kPa/min
<h3>How to calculate dP/dt in an adiabatic expansion?</h3>
An adiabatic process is a process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression
Given b=1.5, P=7 kPa, V=110 cm³, and dV/dt=40 cm³/min
PVᵇ = C
Taking logs of both sides gives:
ln P + b ln V = ln C
Taking partial derivatives gives:
Substitutituting the values b, P, V and dV/dt into the derivative above:
1/7 x dP/dt + 1.5/110 x 40 = 0
1/7 x dP/dt + 6/11 = 0
1/7 x dP/dt = - 6/11
dP/dt = - 6/11 x 7
dP/dt = -42/11 kPa/min
Therefore, the value of dP/dt is -42/11 kPa/min
Learn more about adiabatic expansion on:
brainly.com/question/6966596
#SPJ1
Answer:
54
Step-by-step explanation:
I'm assuming the equation looks like this: 15x - 6
15*4 - 6
60 - 6
54
OR if it's just x-6 then the answer is -2
Hope this is what you were asking, have a nice day! :)
Answer:
let cow=x and chickens =y
cows have 4 legs, 4x. chickens have 2 legs, 2y
IQR = 40
1) Put the numbers in order: 40, 45, 50, 60, 60, 75, 90, 90, 120
2) Find the median: Median is 60 (the 2nd one)
3) Place parentheses around the numbers above and below the median. For easy identification of Q1 and Q3. (40, 45, 50, 60,) 60, (75, 90, 90, 120)
4) Find the Q1 and Q3. Q1 = median of the lower half of the data; Q3 = median of the higher half of the data. Q1 and Q3 have even sets so its median cannot be defined.
5) Had both sets contain odd sets, the median of Q1 is subtracted from the median of Q3 to get the IQR.
We can then use the Alternative definition of IQR.
IQR is the difference between the largest and smallest values in the middle 50% of a set data.
40, 45, 50, 60, 60, 75, 90, 90, 120
Middle 50% is 50, 60, 60, 75, 90; IQR = Largest value - smallest value;
IQR = 90 - 50 = 40
