The three physical forms are:
Solid, liquid, or gas.
The answer is Decibels. <span />
Answer:
Sometimes may cause involuntary responses like twitching
Explanation:
Answer: A.AB
Explanation:
This Velocity vs Time graph shows the acceleration of a body or object, since acceleration is the variation of velocity in time.
As we can see in the attached image, the graph can be divided in four segments:
OA: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a positive slope, hence we are dealing with a positive uniform acceleration.
AB: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be positive. This means the <u>acceleration is nonuniform and positive.</u>
BC: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be negative. This means the acceleration is nonuniform and negative.
CD: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a negative slope, hence we are dealing with a negative uniform acceleration.
From all these segments, the only one that fulfils the nonuniform positive acceleration condition is option A:
Segment AB
Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
![T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}](https://tex.z-dn.net/?f=T_2%20%3D%20T_1%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%5E%7B%5Cfrac%7Bk-1%20%7D%7Bk%7D)
k = 1.4
![T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K](https://tex.z-dn.net/?f=T_2%20%3D%20T_1%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%5E%7B%5Cfrac%7Bk-1%20%7D%7Bk%7D%7D%5C%5C%5C%5CT_2%20%3D%201200%28%5Cfrac%7B80%7D%7B150%7D%29%5E%7B%5Cfrac%7B1.4-1%20%7D%7B1.4%7D%7D%5C%5C%5C%5CT_2%20%3D%201002.714K)
Work done is given as;
![W = \frac{1}{2} *m*(v_i^2 - v_e^2)](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2Am%2A%28v_i%5E2%20-%20v_e%5E2%29)
inlet velocity is negligible;
![v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s](https://tex.z-dn.net/?f=v_e%20%3D%20%5Csqrt%7B%5Cfrac%7B2W%7D%7Bm%7D%20%7D%20%3D%20%5Csqrt%7B2%2AC_p%28T_1-T_2%29%7D%20%5C%5C%5C%5Cv_e%20%3D%20%5Csqrt%7B2%2A1004%281200-1002.714%29%7D%5C%5C%5C%5Cv_e%20%3D%20%5Csqrt%7B396150.288%7D%20%5C%5C%5C%5Cv_e%20%3D%20629.41%20%20%5C%20m%2Fs)
Therefore, the exit velocity is 629.41 m/s