Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure =
= 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
Answer:
mass of the air = 14.62kg
Workdone = 415.88 kJ
Heat transfer = 415.88 kJ
Explanation:
Please find the attached files for the solution
Answer: Doctrine of ratification
Explanation:Doctrine of ratification tries to show that a person who is taking so much a time to complain has indicated that he agrees even if he doesn't without a written consent. This is to eliminate undue waste of time in business, legal an other proceedings requiring consent of both parties.
Doctrine of ratification can either be implied or expressed
Implied ratification is the type of ratification where a persons actions or body language can be seen that he has accepted.
Express ratification is a ratification where a person intentionally accept by showing either through written or verbally.
Answer:
t = 2244.3 sec
Explanation:
calculate the thermal diffusivity


Temperature at 28 mm distance after t time = = 50 degree C
we know that

![\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]](https://tex.z-dn.net/?f=%5Cfrac%7B%2050%20-25%7D%7B300-25%7D%20%3D%20erf%20%5B%5Cfrac%7B28%5Ctimes%2010%5E%7B-3%7D%7D%7B2%5Csqrt%7B1.34%5Ctimes%2010%5E%7B-5%7D%5Ctimes%20t%7D%7D%5D)

from gaussian error function table , similarity variable w calculated as
erf w = 0.909
it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have
w = 0.08073
![erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]](https://tex.z-dn.net/?f=erf%200.08073%20%3D%20erf%5B%5Cfrac%7B3.8245%7D%7B%5Csqrt%7Bt%7D%7D%5D)

solving fot t we get
t = 2244.3 sec
Answer:
a)
(Ω-m)^{-1}
b) Resistance = 121.4 Ω
Explanation:
given data:
diameter is 7.0 mm
length 57 mm
current I = 0.25 A
voltage v = 24 v
distance between the probes is 45 mm
electrical conductivity is given as

![\sigma = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%5Cfrac%7B0.25%20%5Ctimes%2045%5Ctimes%2010%5E%7B-3%7D%7D%7B24%20%5Cpi%20%5B%5Cfrac%7B7%20%5Ctimes%2010%5E%7B-3%7D%7D%7B2%7D%5D%5E2%7D)
(Ω-m)^{-1}[/tex]
b)


![= \frac{57 \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B57%20%20%5Ctimes%2010%5E%7B-3%7D%7D%7B12.2%20%5Ctimes%20%5Cpi%20%5B%5Cfrac%7B7%20%5Ctimes%2010%5E%7B-3%7D%7D%7B2%7D%5D%5E2%7D)
Resistance = 121.4 Ω