The answer would be A, as B refers to conduction and C and D refer to radiation. Convection is the transfer of different temperature currents, i.e, A
We can find the momentum of the rock by using De Broglie's relationship:
![p= \frac{h}{\lambda}](https://tex.z-dn.net/?f=p%3D%20%5Cfrac%7Bh%7D%7B%5Clambda%7D%20)
where
p is the momentum
h is the Planck constant
![\lambda](https://tex.z-dn.net/?f=%5Clambda)
is the De Broglie's wavelength
By using
![\lambda=3.32 \cdot 10^{-34} m](https://tex.z-dn.net/?f=%5Clambda%3D3.32%20%5Ccdot%2010%5E%7B-34%7D%20m)
, we find
![p= \frac{6.6 \cdot 10^{-34} Js}{3.32 \cdot 10^{-34} m}=1.99 kg m/s](https://tex.z-dn.net/?f=p%3D%20%5Cfrac%7B6.6%20%5Ccdot%2010%5E%7B-34%7D%20Js%7D%7B3.32%20%5Ccdot%2010%5E%7B-34%7D%20m%7D%3D1.99%20kg%20m%2Fs%20)
The momentum of the rock is
![p=mv](https://tex.z-dn.net/?f=p%3Dmv)
where
![m=50 g=0.05 kg](https://tex.z-dn.net/?f=m%3D50%20g%3D0.05%20kg)
is the mass and v is its velocity. Rearranging the equation, we find the speed of the rock:
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)
Answer:
Blue Lighting
Explanation:
In order to make red look black, you must use blue light. The blue would be absorbed and there would be no red light to reflect.