Answer:
A. K = 0.546 eV
B. cooper and iron will not emit electrons
Explanation:
A. This is a problem about photoelectric effect. Then you have the following equation:
(1)
K: kinetic energy of the ejected electron
Ф: Work function of the metal = 2.48eV
h: Planck constant = 4.136*10^{-15} eV.s
λ: wavelength of light = 410nm - 750nm
c: speed of light = 3*10^8 m/s
As you can see in the equation (1), higher the wavelength, lower the kinetic energy. Then, the maximum kinetic energy is obtained with the lower wavelength (410nm). Thus, you replace the values of all variables :
![K=(4.136*10^{-15}eV)\frac{3*10^8m/s}{410*10^{-9}m}-2.48eV\\\\K=0.546eV](https://tex.z-dn.net/?f=K%3D%284.136%2A10%5E%7B-15%7DeV%29%5Cfrac%7B3%2A10%5E8m%2Fs%7D%7B410%2A10%5E%7B-9%7Dm%7D-2.48eV%5C%5C%5C%5CK%3D0.546eV)
B. First you calculate the energy of the photon with wavelengths of 410nm and 750nm
![E_1=(4.136*10^{-15}eV)\frac{3*10^{8}m/s}{410*10^{-9}m}=3.02eV\\\\E_2=(4.13610^{-15}eV)\frac{3*10^{8}m/s}{750*10^{-9}m}=1.6544eV](https://tex.z-dn.net/?f=E_1%3D%284.136%2A10%5E%7B-15%7DeV%29%5Cfrac%7B3%2A10%5E%7B8%7Dm%2Fs%7D%7B410%2A10%5E%7B-9%7Dm%7D%3D3.02eV%5C%5C%5C%5CE_2%3D%284.13610%5E%7B-15%7DeV%29%5Cfrac%7B3%2A10%5E%7B8%7Dm%2Fs%7D%7B750%2A10%5E%7B-9%7Dm%7D%3D1.6544eV)
You compare the energies E1 and E2 with the work functions of the metals and you can conclude:
sodium = 2.3eV < E1
cesium = 2.1 eV < E1
cooper = 4.7eV > E1 (this metal will not emit electrons)
iron = 4.5eV > E1 (this metal will not emit electrons)
Answer:
360 V
Explanation:
We can use the transformer equation:
![\frac{V_p}{N_p}=\frac{V_s}{N_s}](https://tex.z-dn.net/?f=%5Cfrac%7BV_p%7D%7BN_p%7D%3D%5Cfrac%7BV_s%7D%7BN_s%7D)
where
is the primary (input) voltage
is the number of turns in the primary coil
is the secondary (output) voltage
is the number of turns in the secondary coil
In this problem,
![V_p = 120 V\\N_p = 25\\N_s = 75](https://tex.z-dn.net/?f=V_p%20%3D%20120%20V%5C%5CN_p%20%3D%2025%5C%5CN_s%20%3D%2075)
Substituting into the equation, we can find the output voltage:
![V_s = N_s \frac{V_p}{N_p}=(75)\frac{120}{25}=360 V](https://tex.z-dn.net/?f=V_s%20%3D%20N_s%20%5Cfrac%7BV_p%7D%7BN_p%7D%3D%2875%29%5Cfrac%7B120%7D%7B25%7D%3D360%20V)
It depends on the angle of the earth, and our point of view. A full moon would occur if we were to be right in front of it.
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters
</span>
Answer:
one billion to ten billion joules.
Explanation:
To keep a 100-watt light bulb going for one second, one hundred joules of energy will be used.