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Nastasia [14]
2 years ago
12

A gas cylinder holds 0.10 mol of O2 at 150 C and a pressure of 3.0 atm. The gas expands adiabatically until the pressure is halv

ed
Part A
What is the final volume?
Part B
What is the final temperature?
Physics
1 answer:
yKpoI14uk [10]2 years ago
3 0

Answer:

V2 = 1.899*10^-3 m^3

T2 = 347.125 K

Explanation:

Using gas law, we know that

PV = nRT,

Where

V1 = 0.00115743 m^3.

gamma = 1.4

Now, when we solve for final volume, V2 we get

V2 = V1/((P2/P1)^(1/gamma))

V2 = 1.899*10^-3 m^3

Using the same law and method, when we try to solve for the temperature, we find that the final temperature, T2 is

T2 = T1*((V1/V2)^(gamma-1))

T2 = 347.125 K

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Calculate the required rate of return for Climax Inc., assuming that (1) investors expect a 4.0% rate of inflation in the future
Westkost [7]

Answer:

Required rate of return = 18.5 %

Explanation:

given,                            

rate of inflection = 4 %

risk free rate = 3 %                      

market risk premium = 5 %                    

firm has a beta  = 2.30                                  

rate of return has averaged 15.0% over the last 5 years

now,                                                                      

Nominal risk free rate = risk free rate  + inflation

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Required rate of return = Nominal risk free rate + β (RPM)

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5 0
3 years ago
Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let |a| = 22 lb and |b| = 16 lb. Rou
SVEN [57.7K]

Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)  

Answer:

Resultant Force=33.8 lb

Angle=67.2°

Explanation:

Given data

Fa=22 lb

Fb=16 lb

Θ=55⁰

To find

(i) Resultant Force F

(ii)Angle α

Solution

First we need to represent the forces in vector form

\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j

Total Force

F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j

The Resultant Force is given as

|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb

For(ii) angle

We can find the angle bu using tanα=y/x

So

tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}

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Explanation:

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