The efficiency of the device is 30 %
Explanation:
The efficiency of a heat engine is given by:

where
W is the work done by the engine
is the heat in input to the engine
For the device in this problem, we have:
W = 120 J is the work done
is the heat in input
Substituting, we find the efficiency:

which corresponds to an efficiency of 30%.
Learn more about work:
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"As frequency increases, wavelength decreases. Frequency and wavelength are inversely proportional. This basically means that when the wavelength is increased, the frequency decreases and vice versa. Wavelength is described as the distance between a trough to a trough or a crest to a crest."
I'd recommend paraphrasing it tho.
Atmospheric pressure decreases with the increase in altitude.
hope that helps you!
Answer:
4.24nm
0.385eV
Explanation:
Maximum wavelength (λmax) :
λmax = ( hc) /Φ
h = plancks constant = 6.63 * 10^-34
c = speed of light = 3*10^8
1ev = 1.6 * 10^-19
Φ = 2.93eV = 2.93* (1.6*10^-19) = 4.688*10^-19
λmax = [(6.63 * 10^-34) * (3 * 10^8)] / 4.688*10^-19
λmax = 19.89 * 10^-26 / 4.688*10^-19
λmax = 4.242 * 10^-7 m
λmax= 4.24nm
B.)
E = hc / eλ eV
λ = 3.75nm = 3.75 * 10^-7m = 375 *10^-9
E = (6.63 * 10^-34) * (3 * 10^8) / (1.6 * 10^-19) * (375 * 10^-9)
E = 19.89 * 10^-26 / 600 * 10^-28
E = 0.03315 * 10^-26 + 28
E = 0.03315 * 10^2
E = 3.315 eV
Stopping potential : (3.315 eV - 2.93eV) = 0.385eV
f = 10000N
distance moved = 20 m
work done = 10000*20 = 200000