<span>♦The gas may become a liquid as it loses energy♦</span>
Answer:
"Longitudinal wave" is the appropriate answer.
Explanation:
- Generating waves whenever the form of communication being displaced in a similar direction as well as in the reverse way of the wave's designated points, could be determined as Longitudinal waves.
- A wave running the length of something like a Slinky stuffed animal, which expands as well as reduces the spacing across spindles, produces a fine image or graphic.
To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.
In other words the acceleration can be described as
![a = \frac{GM}{r^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BGM%7D%7Br%5E2%7D)
Where
G = Gravitational Universal Constant
M = Mass of Earth
r = Radius of Earth
This equation can be differentiated with respect to the radius of change, that is
![\frac{da}{dr} = -2\frac{GM}{r^3}](https://tex.z-dn.net/?f=%5Cfrac%7Bda%7D%7Bdr%7D%20%3D%20-2%5Cfrac%7BGM%7D%7Br%5E3%7D)
![da = -2\frac{GM}{r^3}dr](https://tex.z-dn.net/?f=da%20%3D%20-2%5Cfrac%7BGM%7D%7Br%5E3%7Ddr)
At the same time since Newton's second law we know that:
![F_w = ma](https://tex.z-dn.net/?f=F_w%20%3D%20ma)
Where,
m = mass
a =Acceleration
From the previous value given for acceleration we have to
![F_W = m (\frac{GM}{r^2} ) = 600N](https://tex.z-dn.net/?f=F_W%20%3D%20m%20%28%5Cfrac%7BGM%7D%7Br%5E2%7D%20%29%20%3D%20600N)
Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:
![dF_W = mda](https://tex.z-dn.net/?f=dF_W%20%3D%20mda)
![dF_W = m(-2\frac{GM}{r^3}dr)](https://tex.z-dn.net/?f=dF_W%20%3D%20m%28-2%5Cfrac%7BGM%7D%7Br%5E3%7Ddr%29)
![dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})](https://tex.z-dn.net/?f=dF_W%20%3D%20-2%28m%5Cfrac%7BGM%7D%7Br%5E2%7D%29%28%5Cfrac%7Bdr%7D%7Br%7D%29)
![dF_W = -2F_W(\frac{dr}{r})](https://tex.z-dn.net/?f=dF_W%20%3D%20-2F_W%28%5Cfrac%7Bdr%7D%7Br%7D%29)
But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:
![dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})](https://tex.z-dn.net/?f=dF_W%20%3D%20-2%28600%29%28%5Cfrac%7B1.6%2A10%5E3%7D%7B6.37%2A10%5E6%7D%29)
![dF_W = -0.3N](https://tex.z-dn.net/?f=dF_W%20%3D%20-0.3N)
Therefore there is a weight loss of 0.3N every kilometer.
B. evaporation
c. condensation
They are opposite processes that involve the same transfer of energy
Here we will the speed of seagull which is v = 9 m/s
this is the speed of seagull when there is no effect of wind on it
now in part a)
if effect of wind is in opposite direction then it travels 6 km in 20 min
so the average speed is given by the ratio of total distance and total time
![v_{avg} = \frac{6000}{20*60}](https://tex.z-dn.net/?f=v_%7Bavg%7D%20%3D%20%5Cfrac%7B6000%7D%7B20%2A60%7D)
![v_{avg} = 5m/s](https://tex.z-dn.net/?f=%20v_%7Bavg%7D%20%3D%205m%2Fs)
now since effect of wind is in opposite direction then we can say
![V_{net} = v_{bird} - v_{wind}](https://tex.z-dn.net/?f=V_%7Bnet%7D%20%3D%20v_%7Bbird%7D%20-%20v_%7Bwind%7D)
![5 = 9 - v_{wind}](https://tex.z-dn.net/?f=5%20%3D%209%20-%20v_%7Bwind%7D)
![v_{wind}= 4 m/s](https://tex.z-dn.net/?f=v_%7Bwind%7D%3D%204%20m%2Fs)
Part b)
now if bird travels in the same direction of wind then we will have
![v_{net}= v_{bird} + v_{wind}](https://tex.z-dn.net/?f=v_%7Bnet%7D%3D%20v_%7Bbird%7D%20%2B%20v_%7Bwind%7D)
![v_{net} = 9 + 4 = 13 m/s](https://tex.z-dn.net/?f=v_%7Bnet%7D%20%3D%209%20%2B%204%20%3D%2013%20m%2Fs)
now we can find the time to go back
![time = \frac{distance}{speed}](https://tex.z-dn.net/?f=time%20%3D%20%5Cfrac%7Bdistance%7D%7Bspeed%7D)
![time = \frac{6000}{13}](https://tex.z-dn.net/?f=time%20%3D%20%5Cfrac%7B6000%7D%7B13%7D)
![time = 7.7 minutes](https://tex.z-dn.net/?f=time%20%3D%207.7%20minutes)
Part c)
Total time of round trip when wind is present
![T = t_1 + t_2](https://tex.z-dn.net/?f=T%20%3D%20t_1%20%2B%20t_2)
![T = 20 + 7.7 = 27.7 min](https://tex.z-dn.net/?f=T%20%3D%2020%20%2B%207.7%20%3D%2027.7%20min)
now when there is no wind total time is given by
![T = \frac{6000}{9} + \frac{6000}{9}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B6000%7D%7B9%7D%20%2B%20%5Cfrac%7B6000%7D%7B9%7D)
![T = 22.22 min](https://tex.z-dn.net/?f=T%20%3D%2022.22%20min)
So due to wind time will be more