The energy of a single photon is given by
![E=hf](https://tex.z-dn.net/?f=E%3Dhf)
where
![h=6.6 \cdot 10^{-34} Js](https://tex.z-dn.net/?f=h%3D6.6%20%5Ccdot%2010%5E%7B-34%7D%20Js)
is the Planck constant
f is the frequency of the wave (of the photon)
In our problem, the radio wave has a frequency of
![f=4.40 \cdot 10^8 Hz](https://tex.z-dn.net/?f=f%3D4.40%20%5Ccdot%2010%5E8%20Hz)
, so if we put this value into the previous formula, we can find the energy of a single photon of this electromagnetic wave:
Answer:
The correct Option is (a) please see attachment containing complete question.
Explanation:
1) The motion is basic to and fro motion of an object.
2) Considering the motion as Simple Harmonic Motion (SHM), the graph tells us that the pendulum appears to be moving to the left and slowing down being acted upon positive acceleration.
3) Reaching the left extreme, it stops i.e. V = 0 m/s, and then starts moving to the right due to the component of gravity (g), speeding up.
Cause bubbles are magical
A. Should be the correct answer:) hope this helped! :)
Answer:
![(a)F_{max}=588N\\(b)acceleration=1.96m/s^2](https://tex.z-dn.net/?f=%28a%29F_%7Bmax%7D%3D588N%5C%5C%28b%29acceleration%3D1.96m%2Fs%5E2)
Explanation:
Given data
Mass m=120.0 kg
Coefficient of static friction ![u_{k}=0.500](https://tex.z-dn.net/?f=u_%7Bk%7D%3D0.500)
The coefficient of sliding friction ![u_{d}=0.300](https://tex.z-dn.net/?f=u_%7Bd%7D%3D0.300)
For Part (a) Maximum force
According to Newtons second law the net force aced on the body is given by
![F_{net}=ma=F_{max}-f_{stat}=0\\so\\F_{max}=f_{stat}](https://tex.z-dn.net/?f=F_%7Bnet%7D%3Dma%3DF_%7Bmax%7D-f_%7Bstat%7D%3D0%5C%5Cso%5C%5CF_%7Bmax%7D%3Df_%7Bstat%7D)
The friction force is given by
![f_{stat}=u_{stat}*N\\f_{stat}=0.5*(9.8*120)\\f_{stat}=588N](https://tex.z-dn.net/?f=f_%7Bstat%7D%3Du_%7Bstat%7D%2AN%5C%5Cf_%7Bstat%7D%3D0.5%2A%289.8%2A120%29%5C%5Cf_%7Bstat%7D%3D588N)
Conclude that
![F_{max}=f_{stat}=588N](https://tex.z-dn.net/?f=F_%7Bmax%7D%3Df_%7Bstat%7D%3D588N)
For Part (b) Acceleration
The acceleration due to dynamic friction is given by:
![ma=F_{max}-f_{dyn}](https://tex.z-dn.net/?f=ma%3DF_%7Bmax%7D-f_%7Bdyn%7D)
The dynamic friction is given by:
![f_{dyn}=u_{dyn}*N\\f_{dyn}=0.300*(9.8*120)\\f_{dyn}=353N](https://tex.z-dn.net/?f=f_%7Bdyn%7D%3Du_%7Bdyn%7D%2AN%5C%5Cf_%7Bdyn%7D%3D0.300%2A%289.8%2A120%29%5C%5Cf_%7Bdyn%7D%3D353N)
So the acceleration given by
![a=\frac{F_{max}-f_{dyn}}{m}\\ a=\frac{588N-353N}{120kg}\\ a=1.96m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF_%7Bmax%7D-f_%7Bdyn%7D%7D%7Bm%7D%5C%5C%20a%3D%5Cfrac%7B588N-353N%7D%7B120kg%7D%5C%5C%20a%3D1.96m%2Fs%5E2)