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)
Two types of mechanical waves: longitudinal<span> waves and </span>transverse<span> waves; the medium movement differs between the two.
</span>In a longitudinal wave the medium particle movement is parallel to the direction of wave propagation; example is sound wave in air.
I<span>n a transverse wave the medium particle movement is perpendicular to the direction of wave propagation; example is mechanical wave on a string.
</span><span>
</span>
Answer:
f = pl / (l + p)
Explanation:
1/f = 1/p + 1/l
Find the common denominator of the right hand side.
1/f = l/(pl) + p/(pl)
Add:
1/f = (l + p) / (pl)
Take the inverse of both sides:
f = pl / (l + p)
Answer:
Velocity of truck will be 20.287 m /sec
Explanation:
We have given mass of the truck m = 4000 kg
Radius of the turn r = 70 m
Coefficient of friction ![\mu =0.6](https://tex.z-dn.net/?f=%5Cmu%20%3D0.6)
Centripetal force is given ![F=\frac{mv^2}{r}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7Bmv%5E2%7D%7Br%7D)
And frictional force is equal to ![F_{frictional}=\mu mg](https://tex.z-dn.net/?f=F_%7Bfrictional%7D%3D%5Cmu%20mg)
For body to be move these two forces must be equal
So ![\frac{mv^2}{r}=\mu mg](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E2%7D%7Br%7D%3D%5Cmu%20mg)
![v=\sqrt{\mu rg}=\sqrt{0.6\times 70\times 9.8}=20.287m/sec](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cmu%20rg%7D%3D%5Csqrt%7B0.6%5Ctimes%2070%5Ctimes%209.8%7D%3D20.287m%2Fsec)