Answer:
Explanation:
Calculating the exit temperature for K = 1.4
The value of
is determined via the expression:

where ;
R = universal gas constant = 
k = constant = 1.4


The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :
------ equation(1)
we can rewrite the above equation as :


where:



Thus, the exit temperature = 402.36 K
The exit pressure is determined by using the relation:



Therefore, the exit pressure is 17.79 bar
I believe that the answer is C. Hope this Helps:)))
Given :
Liquid is poured into a burrete so that it reads 14cm³.
50 drops were run each of volume 0.1cm³ .
To Find :
The volume of liquid in burrete after 50 drops.
Solution :
Volume of each drop, v = 0.1 cm³.
Initial volume in burrete, V = 14 cm³.
Now, volume left after droping 50 drops are :

Therefore, the volume left in burrete is 9 cm³ .
Answer:
Explanation:
A.
Given:
Vo = 21 m/s
Vf = 0 m/s
Using equation of Motion,
Vf^2 = Vo^2 - 2aS
S = (21^2)/2 × 9.8
= 22.5 m.
B.
Given:
S = 22.5 + 21 mm
= 22.521 m
Vo = 0 m/s
Using the equation of motion,
S = Vo × t + 1/2 × a × t^2
22.521 = 0 + 1/2 × 9.8 × t^2
t^2 = (2 × 22.521)/9.8
= 4.6
t = 2.14 s
Theories
If it’s wrong oh well because I just guest buh glad to help :)