Answer:
2 seconds
Explanation:
f = Frequency of yell = 919 Hz
= Shifted frequency = 55.9 Hz
v = Speed of sound in air = 342 m/s
= Velocity of friend
a = Acceleration due to gravity = 9.81 m/s²
From the Doppler shift formula we have
![\dfrac{f+\Delta f}{f}=\dfrac{v}{v-v_r}\\\Rightarrow v_r=v-\dfrac{vf}{f+\Delta f}\\\Rightarrow v_r=342-\dfrac{342\times 919}{919+55.9}\\\Rightarrow v_r=19.61\ m/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bf%2B%5CDelta%20f%7D%7Bf%7D%3D%5Cdfrac%7Bv%7D%7Bv-v_r%7D%5C%5C%5CRightarrow%20v_r%3Dv-%5Cdfrac%7Bvf%7D%7Bf%2B%5CDelta%20f%7D%5C%5C%5CRightarrow%20v_r%3D342-%5Cdfrac%7B342%5Ctimes%20919%7D%7B919%2B55.9%7D%5C%5C%5CRightarrow%20v_r%3D19.61%5C%20m%2Fs)
The velocity of the my friend is 19.61 m/s
![v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{19.61-0}{9.8}\\\Rightarrow t=2\ s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7Bv-u%7D%7Ba%7D%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B19.61-0%7D%7B9.8%7D%5C%5C%5CRightarrow%20t%3D2%5C%20s)
The time my friend is in the air is 2 seconds
Answer:
10 seconds
Explanation:
We can calculate the average speed by finding the total distance traveled divided by the elapsed time: Average speed=–s=Total distance Elapsed time
Answer: 0j
Explanation:
At that point potential energy is zero and kinetic energy is maximum.. P. E=mgh=0
Answer:
The answer is A
Explanation:
A large risk of tailgating is the collision avoidance time being much less than the driver reaction time. Driving instructors advocate that drivers always use the "two-second rule" regardless of speed or the type of road. During adverse weather, downhill slopes, or hazardous conditions such as black ice, it is important to maintain an even greater distance.
Answer:
Yes, it can can be completed adiabatically
Explanation:
To solve the problem we will resort to the theory of thermodynamics,
It is necessary to develop this problem to resort to the A-11E tables in English Units for R134a (since the problem requires it, if it were SI just to change to that table)
State 1 indicates that the refrigerant is at 60 ° F,
In the first table (attached image of the value taken) the value of the entropy is
![S_f=S_1= 0.06675Btu/lbm.R](https://tex.z-dn.net/?f=S_f%3DS_1%3D%200.06675Btu%2Flbm.R)
For State 2 the refrigerant is at 50% quality and at a pressure of ![140lbf / in ^ 2](https://tex.z-dn.net/?f=140lbf%20%2F%20in%20%5E%202)
In table 2 of the refrigerant (for the pressure values) we perform the reading and we have to
![Sf= 0.09214](https://tex.z-dn.net/?f=Sf%3D%200.09214)
![Sg=0.21879](https://tex.z-dn.net/?f=Sg%3D0.21879)
We know that,
![S_2 = Sf +X_{quality}(S_g-S_f)](https://tex.z-dn.net/?f=S_2%20%3D%20Sf%20%2BX_%7Bquality%7D%28S_g-S_f%29)
![S_2 = 0.09214+0.5(0.21979-0.09214)](https://tex.z-dn.net/?f=S_2%20%3D%200.09214%2B0.5%280.21979-0.09214%29)
![S_2 = 0.155965BTU/Lb.R](https://tex.z-dn.net/?f=S_2%20%3D%200.155965BTU%2FLb.R)
The change in enthalpy would be given by
![\Delta S = S_2-S_1 = 0.155965BTU/Lb.R- 0.06675Btu/lbm.R](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%20S_2-S_1%20%3D%200.155965BTU%2FLb.R-%200.06675Btu%2Flbm.R)
![\Delta S = 0.089215Btu/lbm.R](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%200.089215Btu%2Flbm.R)
<em>The change in enthalpy is positive, so the process can be completed adiabatically</em>