Question

A: What is the resistance of 6.1 mi of copper wire with a diameter of 0.58 mm ?

Express your answer using two significant figures. (kΩ)

B: Monochromatic light passes through two slits separated by a distance of 0.0340 mm . If the angle to the third maximum above the central fringe is 3.81 ∘ , what is the wavelength of the light (nm)?

C: An AM radio station operating at a frequency of 825 kHz radiates 300 kW of power from its antenna. How many photons are emitted by the antenna every second?

Answer 1

A)

R = rho*L/A

R = 1.68*10^(-8)*6.1*1609.34/((22/7)*(0.29*10^(-3))^2) = 623.97 ohm

B)

Given that
Monochromatic light passes through two slits separated by adistance of ( 2d) = 0.0340 mm.

=0.034*10-3m

The angle to the third maximum above the central fringe is(θ) = 3.81 degree

Then the wavelength of the light is gvien by

sin(theta) = n*lambda/2d

Then the wavelength of the light is

lembda = 2dsin(theta)/n where n =3

Now lambda = (0.034*10^(-3))*sin(3.81)/3

lambda =7.53*10^(-7) m = 753 nm

C)

Energy liberated at given frequency by each photon:

E = hf

E= (6.626*10^(-34))(825000 hz)

E = 5.46*10^(-28) J/photons

number of photons = (330000 J/s) / (5.46*10^(-28) J/photons)

= 6.044*10^32 photons / sec