Answer:
Part a)
![a_t = 0.423 m/s^2](https://tex.z-dn.net/?f=a_t%20%3D%200.423%20m%2Fs%5E2)
Part b)
![a_c = 2113 m/s^2](https://tex.z-dn.net/?f=a_c%20%3D%202113%20m%2Fs%5E2)
Part c)
![d = 80 m](https://tex.z-dn.net/?f=d%20%3D%2080%20m)
Explanation:
Part a)
as we know that angular acceleration of the wheel is given as
![\alpha = 13.2 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%2013.2%20rad%2Fs%5E2)
now the radius of the wheel is given as
R = 3.21 cm
so the tangential acceleration is given as
![a_t = R\alpha](https://tex.z-dn.net/?f=a_t%20%3D%20R%5Calpha)
![a_t = (0.0321)(13.2)](https://tex.z-dn.net/?f=a_t%20%3D%20%280.0321%29%2813.2%29)
![a_t = 0.423 m/s^2](https://tex.z-dn.net/?f=a_t%20%3D%200.423%20m%2Fs%5E2)
Part b)
frequency of the wheel at maximum speed is given as
![f = 2450 rev/min](https://tex.z-dn.net/?f=f%20%3D%202450%20rev%2Fmin)
![f = \frac{2450}{60} = 40.8 rev/s](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B2450%7D%7B60%7D%20%3D%2040.8%20rev%2Fs)
now we know that
![\omega = 2\pi f = 2\pi(40.8) = 256.56 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20f%20%3D%202%5Cpi%2840.8%29%20%3D%20256.56%20rad%2Fs)
now radial acceleration is given as
![a_c = \omega^2 r](https://tex.z-dn.net/?f=a_c%20%3D%20%5Comega%5E2%20r)
![a_c = (256.56)^2(0.0321) = 2113 m/s^2](https://tex.z-dn.net/?f=a_c%20%3D%20%28256.56%29%5E2%280.0321%29%20%3D%202113%20m%2Fs%5E2)
Part c)
total angular displacement of the point on rim is given as
![\Delta \theta = \omega_0 t + \frac{1}{2}\alpha t^2](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%20%5Comega_0%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2)
here we know that
![\omega = \omega_0 + \alpha t](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Comega_0%20%2B%20%5Calpha%20t)
![256.56 = 0 + 13.2 t](https://tex.z-dn.net/?f=256.56%20%3D%200%20%2B%2013.2%20t)
![t = 19.4 s](https://tex.z-dn.net/?f=t%20%3D%2019.4%20s)
now angular displacement will be
![\Delta \theta = 0 + \frac{1}{2}(13.2)(19.4)^2](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%2813.2%29%2819.4%29%5E2)
![\Delta \theta = 2493.3 rad](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%202493.3%20rad)
now the distance moved by the point on the rim is given as
![d = R\theta](https://tex.z-dn.net/?f=d%20%3D%20R%5Ctheta)
![d = (0.0321)(2493.3)](https://tex.z-dn.net/?f=d%20%3D%20%280.0321%29%282493.3%29)
![d = 80 m](https://tex.z-dn.net/?f=d%20%3D%2080%20m)
You have a few choices:
-- Take it out of Earth's atmosphere.
-- Put it in a tank where you pumped all the air out of it.
You can put whatever you want back in the tank, just
as long as there's no oxygen in there.
-- Besides oxygen, rusting also needs water. The humidity
(water vapor) in the air is enough to do it. So if you can
keep the air totally, totally dry around the iron, then the
oxygen alone won't make it rust.
-- All these choices do the same thing: Make sure that
oxygen and water vapor can't reach the iron. The easy
way to do that is to paint or spray something onto the
surface of the iron that keeps the air away from it.
There is no picture to reference from so I’m going to take a wild guess and say A
Answer:
to a warm front. Remember to include all data collected on warm fron … ... Remember to include all data collected on warm fronts in this activity to support your answer (examples: interaction of air masses, air pressure, cloud cover, temperature behind/ahead of front, wind direction, precipitation, etc. 1
Explanation: