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.
Average speed = (total distance covered) / (total time to cover the distance) .
Total distance = (80 + 50 + 40) = 170 km
Total time = (1 + 0.5 + 0.5) = 2 hours
Average speed = (170 km) / (2 hrs) = 85 km/hr .
Answer:
8.5 Ω
Explanation:
La resistencia de un material es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.
La fórmula de la resistencia (R) viene dada por:
R = ρL/A
Donde ρ es la resistividad del material, L es la longitud del material y A es el área de la sección transversal del material.
Dado que:
L = 1 km = 1000 m, A = 2 mm² = 2 * 10⁻⁶ m², ρ (cobre) = 1.7 * 10⁻⁸ Ωm
Sustituyendo da:
R = 1,7 * 10⁻⁸ * 1000/2 * 10⁻⁶
R = 8.5 Ω
Explanation:
We have,
Surface area, 
The current varies wrt time t as :
(a) At t = 2 seconds, electrical charge is given by :
(b) Current is given by :
Instantaneous current at t = 1 s is,
(c) Current is, 
Current density is given by electric current per unit area.
Therefore, it is the required explanation.
Answer:
The magnitude of the induced electric field at a point 2.5 cm from the axis of the solenoid is 8.8 x 10⁻⁵ V/m
Explanation:
given information:
radius, r = 2.0 cm
N = 700 turns/m
decreasing rate, dI/dt = 9.0 A/s
the magnitude of the induced electric field at a point 2.5 cm (r = 2.5 cm = 0.025 m) from the axis of the solenoid?
the magnetic field at the center of solenoid
B = μ₀nI
where
B = magnetic field (T)
μ₀ = permeability (1.26× 10⁻⁶ T.m/A)
n = the number turn per unit length (turn/m)
I = current (A)
dB/dt = μ₀n dI/dt (1)
now we calculate the induced electric field by using
E = 
= 2E/r (2)
where
E = the induced electric field (V/m)
we substitute the firs and second equation, thus
dB/dt = μ₀n dI/dt
2E/r = μ₀n dI/dt
E = (1/2) r μ₀n dI/dt
= (1/2) (0.025) (1.26× 10⁻⁶) (700) (8)
= 8.8 x 10⁻⁵ V/m
