Answer:
![t=45.7s](https://tex.z-dn.net/?f=t%3D45.7s)
![\alpha =116revolutions](https://tex.z-dn.net/?f=%5Calpha%20%3D116revolutions)
Explanation:
Since we have given values of ω₀=32.o rad/s ,ω=0 and α=-0.700 rad/s² to find t we use below equation
![w=w_{o}+at\\ 0=(32.0rad/s)+(-0.700rad/s^{2} )t\\t=\frac{-32.0}{-0.700} \\t=45.7s](https://tex.z-dn.net/?f=w%3Dw_%7Bo%7D%2Bat%5C%5C%20%200%3D%2832.0rad%2Fs%29%2B%28-0.700rad%2Fs%5E%7B2%7D%20%29t%5C%5Ct%3D%5Cfrac%7B-32.0%7D%7B-0.700%7D%20%5C%5Ct%3D45.7s)
To find revolutions we use below equation
![w^{2}=w_{o}^{2}+2a\alpha](https://tex.z-dn.net/?f=w%5E%7B2%7D%3Dw_%7Bo%7D%5E%7B2%7D%2B2a%5Calpha)
Substitute the given values to find revolutions α
So
![0=(32.0rad/s)^{2}+2(-0.700rad/s^{2} )\alpha \\\alpha =\frac{(-32.0rad/s)^{2}}{2(-0.700rad/s^{2} )} \\\alpha =731rad](https://tex.z-dn.net/?f=0%3D%2832.0rad%2Fs%29%5E%7B2%7D%2B2%28-0.700rad%2Fs%5E%7B2%7D%20%29%5Calpha%20%20%5C%5C%5Calpha%20%3D%5Cfrac%7B%28-32.0rad%2Fs%29%5E%7B2%7D%7D%7B2%28-0.700rad%2Fs%5E%7B2%7D%20%29%7D%20%5C%5C%5Calpha%20%3D731rad)
To convert rad to rev:
![\alpha =(731rad)*(\frac{1rev}{2\pi rad} )\\\alpha =116revolutions](https://tex.z-dn.net/?f=%5Calpha%20%3D%28731rad%29%2A%28%5Cfrac%7B1rev%7D%7B2%5Cpi%20rad%7D%20%29%5C%5C%5Calpha%20%3D116revolutions)
Answer: 266.67 s
Explanation:
Average velocity is the rate of change of displacement.
Bird flies 2400 m due west at an average velocity of 9.0 m/s.
![\text{Average velocity}=\frac{Displacement}{time}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20velocity%7D%3D%5Cfrac%7BDisplacement%7D%7Btime%7D)
Given, average velocity, v = 9.0 m/s
Displacement, d = 2400 m
![v=\frac{d}{t}\Rightarrow t=\frac{d}{v}\\ t=\frac{2400 m}{9.0,m/s}=266.67s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bd%7D%7Bt%7D%5CRightarrow%20t%3D%5Cfrac%7Bd%7D%7Bv%7D%5C%5C%20t%3D%5Cfrac%7B2400%20m%7D%7B9.0%2Cm%2Fs%7D%3D266.67s)
Thus, bird was in flight for 266.67 s to fly from nest to land in a tree 2400 m due west.
The correct answer is B. hopes this help
The beat frequency (4 beat per second = 4 Hz ) corresponds to the difference between the frequencies of the two waves:
![4 Hz = f_1 - f_2](https://tex.z-dn.net/?f=4%20Hz%20%3D%20f_1%20-%20f_2)
(1)
The frequency of the first wave can be written as
![f_1= \frac{v}{\lambda_1}](https://tex.z-dn.net/?f=f_1%3D%20%5Cfrac%7Bv%7D%7B%5Clambda_1%7D%20)
where v is the speed of sound in the gas and
![\lambda_1 = 0.8 m](https://tex.z-dn.net/?f=%5Clambda_1%20%3D%200.8%20m)
is the wavelength of the first wave, while the frequency of the second wave is
![f_2 = \frac{v}{\lambda_2}](https://tex.z-dn.net/?f=f_2%20%3D%20%20%5Cfrac%7Bv%7D%7B%5Clambda_2%7D%20)
where
![\lambda_2=0.81 m](https://tex.z-dn.net/?f=%5Clambda_2%3D0.81%20m)
is the wavelength of the second wave. If we substitute into the first equation (1), we find
![4 Hz = \frac{v}{\lambda_1}- \frac{v}{\lambda_2} = v (\frac{1}{\lambda_2}- \frac{1}{\lambda_2} )](https://tex.z-dn.net/?f=4%20Hz%20%3D%20%20%5Cfrac%7Bv%7D%7B%5Clambda_1%7D-%20%5Cfrac%7Bv%7D%7B%5Clambda_2%7D%20%3D%20v%20%28%5Cfrac%7B1%7D%7B%5Clambda_2%7D-%20%5Cfrac%7B1%7D%7B%5Clambda_2%7D%20%29)
and if we re-arrange it, we can find the velocity of the sound in the gas:
Answer:
Microwave spectroscopy is the spectroscopy method that employs microwaves, i.e. electromagnetic radiation at GHz frequencies, for the study of matter