Out of the following choices, Newton's Third Law of Motion states that when one objects exerts a force on a second object, the forces are equal in magnitude and opposite in direction.
Answer:
- 8632.1N/C
- 17483.17N/C
Explanation:
Inside electric field magnitude is different of the outside electric field.
- Inside the sphere we have that the electric field is given by:
![E=\frac{Qr}{4\pi \epsilon_oR^3}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BQr%7D%7B4%5Cpi%20%5Cepsilon_oR%5E3%7D)
Where Q is the charge of the sphere, R is its radius and e0 is the dielectric permittivity of vacuum.
By replacing (r=5.00cm=5.00*10^{-3}m, e0=8.85*10^{-12}C^2/Nm^2) we get:
![E(r=4.00cm)=\frac{(3.00*10^{-9}C)(0.04m)}{4\pi (8.85*10^{-12}C^2/Nm^2)(0.05m)^3}=8632.1N/C](https://tex.z-dn.net/?f=E%28r%3D4.00cm%29%3D%5Cfrac%7B%283.00%2A10%5E%7B-9%7DC%29%280.04m%29%7D%7B4%5Cpi%20%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%280.05m%29%5E3%7D%3D8632.1N%2FC)
- Outside the sphere we have the formula:
![E=\frac{Q}{4\pi \epsilom_or^2}\\\\E(r=6.00cm)=\frac{3.00*10^{-9}C}{4\pi (8.85*10^{-12}C^2/Nm^2)(0.06)^2}=17483.17N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BQ%7D%7B4%5Cpi%20%5Cepsilom_or%5E2%7D%5C%5C%5C%5CE%28r%3D6.00cm%29%3D%5Cfrac%7B3.00%2A10%5E%7B-9%7DC%7D%7B4%5Cpi%20%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%280.06%29%5E2%7D%3D17483.17N%2FC)
hope this helps!!
Answer:
2400 kilogram per cubic meter
Explanation:
the density of concrete is a measure of its unit weight.
concrete is a mixture of cement, fine and coarse aggregates, water and sometimes some supplementary materials like fly ash, and various admixtures.
Answer:
Common temporary magnets include nails and paperclips, which can be picked up or moved by a strong magnet. Another type of temporary magnet is an electromagnet which only retains magnetism when an electrical current is running through it.
Explanation:
The energy levels of the hydrogen atom are given by
![E_n = -13.6 \frac{1}{n^2} [eV]](https://tex.z-dn.net/?f=E_n%20%3D%20-13.6%20%20%5Cfrac%7B1%7D%7Bn%5E2%7D%20%5BeV%5D%20)
where n is the level number. Let's use this formula to calculate the energy of the levels n=1 (ground state), n=5 and n=6:
- ground state:
![E_1 = -13.6 \frac{1}{1^2} eV=-13.6 eV = -2.18 \cdot 10^{-18} J](https://tex.z-dn.net/?f=E_1%20%3D%20-13.6%20%20%5Cfrac%7B1%7D%7B1%5E2%7D%20eV%3D-13.6%20eV%20%3D%20-2.18%20%5Ccdot%2010%5E%7B-18%7D%20J)
- level n=5:
![E_5 = -13.6 \frac{1}{5^2} eV = -13.6 \frac{1}{25} eV=-0.54 eV =-8.64 \cdot 10^{-20} J](https://tex.z-dn.net/?f=E_5%20%3D%20-13.6%20%20%5Cfrac%7B1%7D%7B5%5E2%7D%20eV%20%3D%20-13.6%20%20%5Cfrac%7B1%7D%7B25%7D%20eV%3D-0.54%20eV%20%3D-8.64%20%5Ccdot%2010%5E%7B-20%7D%20J%20%20)
- level n=6:
![E_6 = -13.6 \frac{1}{6^2}eV= -13.6 \frac{1}{36} eV = -0.38 eV = -6.08 \cdot 10^{-20} J](https://tex.z-dn.net/?f=E_6%20%3D%20-13.6%20%20%5Cfrac%7B1%7D%7B6%5E2%7DeV%3D%20-13.6%20%5Cfrac%7B1%7D%7B36%7D%20eV%20%3D%20-0.38%20eV%20%3D%20%20-6.08%20%5Ccdot%2010%5E%7B-20%7D%20J)
Now that we have the energy for all the levels we are interested in, we can calculate the energy of the emitted photons.
a) In the first transition, the atom goes from n=6 to n=5. The energy of the emitted photon is equal to the energy difference between these two levels:
![E=E_6-E_5 =-6.09 \cdot 10^{-20} J-(-8.65 \cdot 10^{-20}J)=2.56 \cdot 10^{-20}J](https://tex.z-dn.net/?f=E%3DE_6-E_5%20%3D-6.09%20%5Ccdot%2010%5E%7B-20%7D%20J-%28-8.65%20%5Ccdot%2010%5E%7B-20%7DJ%29%3D2.56%20%5Ccdot%2010%5E%7B-20%7DJ)
The energy of a photon is also equal to
![E=hf=h \frac{c}{\lambda}](https://tex.z-dn.net/?f=E%3Dhf%3Dh%20%5Cfrac%7Bc%7D%7B%5Clambda%7D%20)
where
h is the Planck constant
c is the speed of light
f is the photon frequency
is its wavelength
Re-arranging this relationship and using the photon's energy, we find its wavelength:
![\lambda= \frac{hc}{E}= \frac{(6.63 \cdot 10^{-34}Js)(3 \cdot 10^8 m/s)}{2.56 \cdot 10^{-20} J}=7.77 \cdot 10^{-6} m = 7770 nm](https://tex.z-dn.net/?f=%5Clambda%3D%20%5Cfrac%7Bhc%7D%7BE%7D%3D%20%5Cfrac%7B%286.63%20%5Ccdot%2010%5E%7B-34%7DJs%29%283%20%5Ccdot%2010%5E8%20m%2Fs%29%7D%7B2.56%20%5Ccdot%2010%5E%7B-20%7D%20J%7D%3D7.77%20%5Ccdot%2010%5E%7B-6%7D%20m%20%3D%207770%20nm)
b) in the second transition, from n=5 to n=1, the energy of the emitted photon is equal to the difference in energy between the two levels:
![E=E_5 - E_1 = -8.65 \cdot 10^{-20} J -(-2.18 \cdot 10^{-18} J) =2.09 \cdot 10^{-18} J](https://tex.z-dn.net/?f=E%3DE_5%20-%20E_1%20%3D%20-8.65%20%5Ccdot%2010%5E%7B-20%7D%20J%20-%28-2.18%20%5Ccdot%2010%5E%7B-18%7D%20J%29%20%3D2.09%20%5Ccdot%2010%5E%7B-18%7D%20J)
Similarly to part a), the wavelength of the photon is given by:
![\lambda= \frac{hc}{E}= \frac{(6.6 \cdot 10^{-34}Js)(3 \cdot 10^8 m/s)}{2.09 \cdot 10^{-18} J}=9.52 \cdot 10^{-8} m = 95.2 nm](https://tex.z-dn.net/?f=%5Clambda%3D%20%5Cfrac%7Bhc%7D%7BE%7D%3D%20%5Cfrac%7B%286.6%20%5Ccdot%2010%5E%7B-34%7DJs%29%283%20%5Ccdot%2010%5E8%20m%2Fs%29%7D%7B2.09%20%5Ccdot%2010%5E%7B-18%7D%20J%7D%3D9.52%20%5Ccdot%2010%5E%7B-8%7D%20m%20%3D%2095.2%20nm)