Hot combustion gases are accelerated in a 92% efficient
adiabatic nozzle from low velocity to a specified velocity. The exit velocity
and the exit temp are to be determined.
Given:
T1 = 1020 K à
h1 = 1068.89 kJ/kg, Pr1 = 123.4
P1 = 260 kPa
T1 = 747 degrees Celsius
V1 = 80 m/s ->nN = 92% -> P2
= 85 kPa
Solution:
From the isentropic relation,
Pr2<span> = (P2 / P1)PR1 = (85
kPa / 260 kPa) (123.4) = 40.34 = h2s = 783.92 kJ/kg</span>
There is only one inlet and one exit, and thus, m1 =
m2 = m3. We take the nozzle as the system, which is a
control volume since mass crosses the boundary.
h2a = 1068.89 kJ/kg – (((728.2 m/s)2 –
(80 m/s)2) / 2) (1 kJ/kg / 1000 m2/s2) =
806.95 kJ/kg\
From the air table, we read T2a = 786.3 K
Answer:
A. 0.432
B. -1.92
C. 1.44 units/second
D. -3.2 units/second
Explanation:
A. To calculate her x position, we just use the following equation of motion to find the distance traveled:

here s = displacement
t = time (in seconds)
a = acceleration
Solving for the distance, we get:

s = 0.432 m
Since 0.432 meters east is equals to 0.432 meter in the positive x-direction, the x position is also 0.432.
B. Since the skater has a constant v - velocity of -3.2 m/s, (south means negative y axis), the total distance traveled is:
Distance = speed * time = -3.2 * 0.6 = -1.92 m
The answer is -1.92 units in the y-axis.
C. The x velocity component is the final speed in the east direction, which is going to be:


v = 1.44 units/second (in positive x direction)
D. Her y velocity component does not change, since the velocity towards the south is a constant 3.2 m/s
Thus the answer is -3.2 units/second in the y-axis.
Answer:633 m
Explanation:
First we have moved 300 m in North
let say it as point a and its vector is 
after that we have moved 500 m northeast
let say it as point b
therefore position of b with respect to a is
r
Therefore position of b w.r.t to origin is


![r_b=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}](https://tex.z-dn.net/?f=r_b%3D500cos%2845%29%5Chat%7Bi%7D%2B%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B300%5Cright%20%5D%5Chat%7Bj%7D)
after this we moved 400 m
south of east i.e.
below from positive x axis
let say it as c


![r_c=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}+400cos(60)\hat{i}-400sin(60)\hat{j}](https://tex.z-dn.net/?f=r_c%3D500cos%2845%29%5Chat%7Bi%7D%2B%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B300%5Cright%20%5D%5Chat%7Bj%7D%2B400cos%2860%29%5Chat%7Bi%7D-400sin%2860%29%5Chat%7Bj%7D)
![r_c=\left [ 250\sqrt{2}+200\right ]\hat{i}+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]\hat{j}](https://tex.z-dn.net/?f=r_c%3D%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B200%5Cright%20%5D%5Chat%7Bi%7D%2B%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B300-200%5Csqrt%7B3%7D%5Cright%20%5D%5Chat%7Bj%7D)
magnitude is ![\sqrt{\left [ 250\sqrt{2}+200\right ]^2+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B200%5Cright%20%5D%5E2%2B%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B300-200%5Csqrt%7B3%7D%5Cright%20%5D%5E2%7D)
=633.052
for direction

with x -axis
Answer:
Energy gained by the second particle = 12Uo
Explanation:
Given Data;
Resistant force = 12F
Initial kinetic energy = Uo
Calculating the kinetic energy gained, we have;
u = f *r
where f= resistant force = 20F
r = initial kinetic energy = Uo
Therefore,
U = 12 * uo
= 12 Uo
Therefore, energy gained by the second particle = 12Uo
Answer:

Explanation:
We must separate the motion into two parts, the first when the rocket's engines is on and the second when the rocket's engines is off. So, we need to know the height (
) that the rocket reaches while its engine is on and we need to know the distance (
) that it travels while its engine is off.
For solving this we use the kinematic equations:
In the first part we have:

and the final speed is:

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.
So, we have:

The sum of these heights will give us the total height, which is known:

This is the time that its needed in order for the rocket to reach the required altitude.