Answer:
a) 
b)
c)
Explanation:
a) Let's calculate the work done by the rocket until the thrust ends.

But we know the work is equal to change of kinetic energy, so:


b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

At the maximum height the velocity is 0, so v(f) = 0.

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.
Using the same equation of part b)


The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.
I hope it helps you!
Known :
D = 12 in = 1 ft
L = 850 ft
Q = 5.6 cfs
hA = 750 ft
hB = 765 ft
PA = 85 psi = 12240 lb/ft²
Solution :
A = πD² / 4 = π(1²) / 4
A = 0.785 ft²
<u>Velocity of water :</u>
U = Q / A = 5.6 / 0.785
U = 7.134 ft/s
<u>Friction loss due to pipe length :</u>
Re = UD / v = (7.134)(1) / (0.511 × 10^(-5))
Re = 1.4 × 10⁶
(From Moody Chart, We Get f = 0.015)
hf = f(L / d)(U² / 2g) = 0.015(850 / 1)((7.134²) / 2(32.2))
hf = 10 ft
PA + γhA = PB + γhB + γhf
PB = PA + γ(hA - hB - hf)
PB = 12240 + (62.4)(750 - 765 - 10)
PB = 10680 lb/ft²
PB = 74.167 psi
Answer:

Explanation:
Hello,
In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy <em>at constant pressure</em>:

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

We obtain the expression to compute the specific entropy change:

Best regards.
Answer: You're question is very vague... be more specific in the problem, what is it asking you to do?
Explanation: This shouldn't be under "Engineering" you should put it under "Mathematics" for a better result. Sorry for the mix up! Hope this helps! ^^
Automobiles and Airplanes is what I can think of