This is part of a practice final exam for my physics two cla

This is part of a practice final exam for my physics two class. If you could answer all the questions it would be greatly appreciated so i can know if i did it right or not. They aren\'t too hard and most just need one equation and some conversions.

A 4A current is set up in a circuit for 2 minutes by a rechargeable battery with 12 V emf. by how much is the chemical energy of the battery reduced?

three resistors connected in parallel have resistances of 10, 20 and 30 ohms. the sum of the currents passing through is 6 A. what is the potential difference across the set?

a light wave of wavelength 520nm passes from air to water where the index of refraction is 1.32. the wavelength and frequency of light in water are?

a coherent source of monochromatic light of wavelength 600nm shines on a single slit of width 0.3 mm. what is the distance between the first and the second minima in the diffraction pattern of the single slit on a screen 2 meters from the single slit?

Solution

According the given problems,

a) Genarally Ampere= C/s and Volts = J/C

4A in 2 minutes is 4*120 = 480 C

480 C at 12V is 5760 Joules

b)First calcualte the R_eqi

1/R_eqi =1/R₁+1/R₂+1/R₃

R_eqi = 5.4545 ohms

Using amps law

V = I*R_eqi

V = 6* 5.4545

V = 32.7272 V

c) Using the fallowing eqautions

_air/_water = n_water /n_air

520nm /_water = 1.32/1

_water = 520 nm/1.32

_water = 409.1 nm

f = c/_water = 3*10⁸/409.1

f = 7.33 * 10¹⁴Hz

d) Genaral formula for a single slit difration

y = mD/d where where, m = the order of minimum, y = distance on the screen of the minimum from central axis, = wavelength, D = screen-to-slit distance, d = slit width

The distance between the first and the second minima is,

y₂ - y₁ = [2*(600x10)*2 / (0.3*10^-3)] - [1*(600x10)*2 / (0.3*10^-3)]

y₂ - y₁ = 600x10*2 / (0.3*10^-3)

y₂ - y₁ = 4 * 10^-3m

y₂ - y₁ = 4mm