** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **
Ans: The vertical distance = y = M/(ρA)
Explanation:Support the vertical distance = y
Object's density = M/(A*h) (since A*h = volume)
By applying the condition,
(M/(Ah))/ρ = y/h
M/(ρAh) = y/h
y = M/(ρA)
Answer:
7.09683 m
1.20285 s
2.4057 s
11.8 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)
From equation of motion we have
The maximum height above the ground that the ball reaches is 7.09683 m
Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s
The velocity just before it hits the ground is 11.8 m/s
Oxygen is diatomic, so its degree of freedom, (f1)= 5,
also its number of moles, n1= 1
Helium is monoatomic, so its degree of freedom (f2)= 3
and its number of moles given is, n2=2
Now using formula of effective degree of freedom of mixture, (f), we have:
f= (f1n1+f2n2)/(n1+n2)
= (5*1 + 3*2)/ (1+3)
=11/3
Also, from first law of thermodynamics;
U= n Cv. T = nRT(f2)
or, Cv = R. (f/2) (n & T cancel)
We know f=11/6,
substituting the value in above relation, we have:
Cv= R. 11/3*2
= R. 11/6
Also, Cp-Cv = R
or, Cp- R.(11/6)= R
or, Cp= R(11/6 )+1
= 17/6 R
Therefore, Cp/Cv = 17/11
