To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.
For this purpose we have that the constructive interference in waves can be expressed under the function

Where
a = Width of the slit
d = Distance of slit to screen
m = Number of order which represent the number of repetition of the spectrum
Angle between incident rays and scatter planes
At the same time the distance on the screen from the central point, would be

Where y = Represents the distance on the screen from the central point
PART A ) From the previous equation if we arrange to find the angle we have that



PART B) Equation both equations we have


Re-arrange to find a,


First, we need to find the number of protons, which is the total mass divided by the mass of one proton:

protons
Then, the total charge is the number of protons times the charge of a single proton:
64 miles/hour
Therefore 1/64 hours/mile
68 miles * 1/64 hours/mile (notice how miles cancels out)
Therefore the answer is 68/64 hours = 1.0625 hours = 1 hour 3min and 45sec.
Answer: a) 139.4 μV; b) 129.6 μV
Explanation: In order to solve this problem we have to use the Ohm law given by:
V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.
Then we have:
for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω
and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω
Finalle we calculate the potential difference (V) for both wires:
Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V
V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V