Answer:
Option e) 320 s
Explanation:
Here, distance = 3.0 km = 3000 m
The velocity of boat when it is going upstream;
Upstream velocity = velocity of boat in still water - velocity of river flow
So, Upstream velocity ![=20m/s-5m/s=15m/s](https://tex.z-dn.net/?f=%3D20m%2Fs-5m%2Fs%3D15m%2Fs)
So,Time to go upstream
![(t_{1}) =\frac{Distance}{Velocity}=\frac{3000m}{15m/s} =200 s](https://tex.z-dn.net/?f=%28t_%7B1%7D%29%20%3D%5Cfrac%7BDistance%7D%7BVelocity%7D%3D%5Cfrac%7B3000m%7D%7B15m%2Fs%7D%20%3D200%20s)
The velocity of boat when it is going downstream;
Downstream velocity = velocity of boat in still water + velocity of river flow
So, Downstream velocity ![=20m/s+5m/s=25m/s](https://tex.z-dn.net/?f=%3D20m%2Fs%2B5m%2Fs%3D25m%2Fs)
So,Time to go downstream
![(t_{2}) =\frac{Distance}{Velocity}=\frac{3000m}{25m/s} =120 s](https://tex.z-dn.net/?f=%28t_%7B2%7D%29%20%3D%5Cfrac%7BDistance%7D%7BVelocity%7D%3D%5Cfrac%7B3000m%7D%7B25m%2Fs%7D%20%3D120%20s)
So, total time (t) = ![t_{1}+t_{2}=200s+120 s=320s](https://tex.z-dn.net/?f=t_%7B1%7D%2Bt_%7B2%7D%3D200s%2B120%20s%3D320s)
Option E is the correct answer.
Answer:
a) 4.04*10^-12m
b) 0.0209nm
c) 0.253MeV
Explanation:
The formula for Compton's scattering is given by:
![\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)](https://tex.z-dn.net/?f=%5CDelta%20%5Clambda%3D%5Clambda_f-%5Clambda_i%3D%5Cfrac%7Bh%7D%7Bm_oc%7D%281-cos%5Ctheta%29)
where h is the Planck's constant, m is the mass of the electron and c is the speed of light.
a) by replacing in the formula you obtain the Compton shift:
![\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m](https://tex.z-dn.net/?f=%5CDelta%20%5Clambda%3D%5Cfrac%7B6.62%2A10%5E%7B-34%7DJs%7D%7B%289.1%2A10%5E%7B-31%7Dkg%29%283%2A10%5E8m%2Fs%29%7D%281-cos132%5C%C2%B0%29%3D4.04%2A10%5E%7B-12%7Dm)
b) The change in photon energy is given by:
![\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm](https://tex.z-dn.net/?f=%5CDelta%20E%3DE_f-E_i%3Dh%5Cfrac%7Bc%7D%7B%5Clambda_f%7D-h%5Cfrac%7Bc%7D%7B%5Clambda_i%7D%3Dhc%28%5Cfrac%7B1%7D%7B%5Clambda_f%7D-%5Cfrac%7B1%7D%7B%5Clambda_i%7D%29%5C%5C%5C%5C%5Clambda_f%3D4.04%2A10%5E%7B-12%7Dm%20%2B%5Clambda_i%3D4.04%2A10%5E%7B-12%7Dm%2B%280.0169%2A10%5E%7B-9%7Dm%29%3D2.09%2A10%5E%7B-11%7Dm%3D0.0209nm)
c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.
![P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bh%7D%7B%5Clambda_e%7D%3D%5Cfrac%7B6.62%2A10%5E%7B-34%7DJs%7D%7B2.43%2A10%5E%7B-12%7Dm%7D%3D2.72%2A10%5E%7B-22%7Dkgm%5C%5C)
![E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV](https://tex.z-dn.net/?f=E_e%3D%5Cfrac%7Bp%5E2%7D%7B2m_e%7D%3D%5Cfrac%7B%282.72%2A10%5E%7B-22%7Dkgm%29%5E2%7D%7B2%289.1%2A10%5E%7B-31%7Dkg%29%7D%3D4.06%2A10%5E%7B-14%7DJ%5C%5C%5C%5C1J%3D6.242%2A10%5E%7B18%7DeV%5C%5C%5C%5CE_e%3D4.06%2A10%5E%7B-14%7D%286.242%2A10%5E%7B18%7DeV%29%3D0.253MeV)
10800 m = 10.8 km should be the answer if I am correct
Calculate the number of neutrons. Now you know that atomic number = number of protons, and mass number = number of protons + number of neutrons. To find the number of neutrons in an element, subtract the atomic number from the mass number.
Answer:
1.62 atm
Explanation:
We can solve the problem by using the ideal gas equation:
![pV=nRT](https://tex.z-dn.net/?f=pV%3DnRT)
where:
p = ? is the pressure of the gas in the tire
V = 8.5 L is the volume of the tire
n = 0.55 mol is the number of moles of the gas
R = 0.0821 atm L / K mol is the gas constant
T = 305 K is the temperature of the gas
By re-arranging the equation and substituting the numbers in, we find:
![p=\frac{nRT}{V}=\frac{(0.55 mol)(0.0821 Latm/Kmol)(305 K)}{8.5 L}=1.62 atm](https://tex.z-dn.net/?f=p%3D%5Cfrac%7BnRT%7D%7BV%7D%3D%5Cfrac%7B%280.55%20mol%29%280.0821%20Latm%2FKmol%29%28305%20K%29%7D%7B8.5%20L%7D%3D1.62%20atm)