Efficiency = Work Output / Work Input
92% = Work Output / 100
0.92 = Work Output / 100
Work Output = 0.92 * 100
Work Output = 92 joules.
Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
I think it 32, but i’m not sure
According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:
Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:
Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:
Therefore, the answer is: 532,346 J.
We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s
