Answer:
P.E = 82320 joules
Explanation:
Given,
The mass of the squirrel, m = 700 Kg
The squirrel leaped to a height, h = 12 m
The acceleration due to gravity, g = 9.8 m/s²
The potential energy of a body is possessed by its height. It is given by the formula
P.E = mgh joules
Substituting the values in the above equation
P.E = 700 Kg x 9.8 m/s² x 12 m
= 82320 joules
The Potential Energy of the squirrel is, P.E = 82320 joules
-9.8 m/s^2 would be right.
he speed of sound can travel faster in hot air than it can in cold air.
Answer:
Explanation:
(a) This problem can be solved by using the formula for Compton effect
![\lambda_{a}-\lambda_{b}=\frac{h}{m_{o}c}(1-cos\phi)](https://tex.z-dn.net/?f=%5Clambda_%7Ba%7D-%5Clambda_%7Bb%7D%3D%5Cfrac%7Bh%7D%7Bm_%7Bo%7Dc%7D%281-cos%5Cphi%29)
where h is the planck constant, Φ is the angle of the scattered photon, m is the mass of the particle (in this case we take an electron) and c is the speed of ligth.
By taking the wavelength of the scattered photon and by replacing we have
![\lambda_{a}=\frac{6.62*10^{-34}m^{2}\frac{kg}{s}}{(9.1*10^{-31}kg)(3*10^{8}\frac{m}{s})}+0.02480*10^{-9}m=2.72*10^{-11}m=0.0272nm](https://tex.z-dn.net/?f=%5Clambda_%7Ba%7D%3D%5Cfrac%7B6.62%2A10%5E%7B-34%7Dm%5E%7B2%7D%5Cfrac%7Bkg%7D%7Bs%7D%7D%7B%289.1%2A10%5E%7B-31%7Dkg%29%283%2A10%5E%7B8%7D%5Cfrac%7Bm%7D%7Bs%7D%29%7D%2B0.02480%2A10%5E%7B-9%7Dm%3D2.72%2A10%5E%7B-11%7Dm%3D0.0272nm)
(b)
![p=\frac{h}{\lambda_{b}}=\frac{6.62*10^{-34}m^{2}\frac{kg}{s}}{0.0248*10^{-9}m}=2.66*10^{-23}kg\frac{m}{s}](https://tex.z-dn.net/?f=p%3D%5Cfrac%7Bh%7D%7B%5Clambda_%7Bb%7D%7D%3D%5Cfrac%7B6.62%2A10%5E%7B-34%7Dm%5E%7B2%7D%5Cfrac%7Bkg%7D%7Bs%7D%7D%7B0.0248%2A10%5E%7B-9%7Dm%7D%3D2.66%2A10%5E%7B-23%7Dkg%5Cfrac%7Bm%7D%7Bs%7D)
(c)
The energy is conserved, hence we have
![E_{i-photon}=E_{k-electron}+E_{s-photon}](https://tex.z-dn.net/?f=E_%7Bi-photon%7D%3DE_%7Bk-electron%7D%2BE_%7Bs-photon%7D)
that is, the sum of the kinetic energy of the scattered electron and the energy of the scattered photon is equal to the energy of the incident photon. By taking Ek we have
![E_{k-e}=E_{i-p}-E_{s-p}=h\frac{c}{\lambda_{b}}-h\frac{c}{\lambda_{a}}\\\\E_{k-e}=(6.62*10^{-34}m^{2}\frac{kg}{s})(3*10^{8}\frac{m}{s})(\frac{1}{0.0248*10^{-9}m}-\frac{1}{0.0272*10^{-9}m})\\\\E_{k-e}=7.06*10^{-16}J=7.06*10^{-16}(6.242*10^{18}eV)=4.4keV](https://tex.z-dn.net/?f=E_%7Bk-e%7D%3DE_%7Bi-p%7D-E_%7Bs-p%7D%3Dh%5Cfrac%7Bc%7D%7B%5Clambda_%7Bb%7D%7D-h%5Cfrac%7Bc%7D%7B%5Clambda_%7Ba%7D%7D%5C%5C%5C%5CE_%7Bk-e%7D%3D%286.62%2A10%5E%7B-34%7Dm%5E%7B2%7D%5Cfrac%7Bkg%7D%7Bs%7D%29%283%2A10%5E%7B8%7D%5Cfrac%7Bm%7D%7Bs%7D%29%28%5Cfrac%7B1%7D%7B0.0248%2A10%5E%7B-9%7Dm%7D-%5Cfrac%7B1%7D%7B0.0272%2A10%5E%7B-9%7Dm%7D%29%5C%5C%5C%5CE_%7Bk-e%7D%3D7.06%2A10%5E%7B-16%7DJ%3D7.06%2A10%5E%7B-16%7D%286.242%2A10%5E%7B18%7DeV%29%3D4.4keV)
(d)
![E_{k-e}=\frac{m_{e}v^{2}}{2}\\\\v=\sqrt{\frac{2E_{k-e}}{m_{e}}}=3.9*10^{7}\frac{m}{s}\\\\p_{e}=m_{e}*v=3.5*10^{-23}kg\frac{m}{s}](https://tex.z-dn.net/?f=E_%7Bk-e%7D%3D%5Cfrac%7Bm_%7Be%7Dv%5E%7B2%7D%7D%7B2%7D%5C%5C%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B2E_%7Bk-e%7D%7D%7Bm_%7Be%7D%7D%7D%3D3.9%2A10%5E%7B7%7D%5Cfrac%7Bm%7D%7Bs%7D%5C%5C%5C%5Cp_%7Be%7D%3Dm_%7Be%7D%2Av%3D3.5%2A10%5E%7B-23%7Dkg%5Cfrac%7Bm%7D%7Bs%7D)
I hope this is useful for you
regards