Explanation:
It is given that,
Length of the rod, l = 14 cm = 0.14 m
Charge on the rod, 
We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m
Electric field at the axis of the rod is given by :
Where
is the linear charge density of the rod,
E = -7493170.57 N/C
or
Negative sign shows that the electric field is acting in inwards direction. Hence, this is the required solution.
If the light of wavelength 700 nm strikes such a photocathode the maximum kinetic energy, in eV, of the emitted electrons is 0.558 eV.
so - $KE_{max} = hc/lembda} work
threshold when KE = 0
hc/lambda = work = 1240/900=1.38 eV
b) Kemax = hc/lambda - work = 1240/640 -1.38=0.558 eV
What is photocathode?
- A photocathode electrolyte interface can be used in a photoelectrolysis cell as the primary light-harvesting junction (in conjunction with an appropriate electrochemical anode) or as an optically complementary photoactive half-cell in a tandem photoelectrode photoelectrolysis cell (Hamnett, 1982; Kocha et al, 1994).
- In the case of the former, the electrode should ideally harvest photon energy across the majority of the solar spectrum in order to achieve the highest energy conversion efficiency possible.
- In the latter case, however, the photocathode may only be active in a specific band of the solar spectrum in order to generate a cathodic photocurrent sufficient to match the current generated in the photoanodic half-cell.
To learn more about Photocathode from the given link:
brainly.com/question/9861585
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Answer:
Explanation:
Intake heat, QH = 100 J
output heat, Qc = 20 J
Work, W = 80 J
TH = 100°C = 373 K
Tc = 10°C = 283 K
TH/ Tc = 373 / 283 = 1.318
QH/Qc = 100 / 20 = 5
for a heat engine, those ratios should be same. so temperature is not correct.
Answer:
Acceleration due to gravity
