Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t
= λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t
= 750 / 2(1.20)
t
= 750 / 2.4
t
= 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t
= λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t
= 750 / 4(1.50)
t
= 750 / 6
t
= 125 nm
Therefore, the minimum thickness be now will be 125 nm
The one fact that needs to be mentioned but isn't given anywhere on or around the graph is: The distance, on the vertical axis, is the distance FROM home. So any point on the graph where the distance is zero ... the point is in the x-axis ... is a point AT home.
Segment D ...
Walking AWAY from home; distance increases as time increases.
Segment B ...
Not walking; distance doesn't change as time increases.
Segment C ...
Walking away from home, but slower than before; distance increases as time increases, but not as fast. Slope is less than segment-D.
Segment A ...
Going home; distance is DEcreasing as time increases. Walking pretty fast ... the slope of the line is steep.
We have that the amount of the charge q is

From the Question we are told that
Force 
Velocity 
Angle 
Magnetic field 
Generally, the equation for Force F is mathematically given by



In conclusion
The amount of the charge q is

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Answer:
t = 23.9nS
Explanation:
given :
Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m
distance d= 1mm=> 0.001
resistor R= 975 ohm
Capacitance can be calculated through the following formula,
C = (ε0 x A )/d
C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001
C = 17.7 x 10^-12 (pico 'p' = 10^-12)
C = 17.7pF
the voltage between two plates is related to time, There we use the following formula of the final voltage
Vc = Vx (1-e^-(t/CR))
75 = 100 x (1-e^-(t/CR))
75/100 = (1-e^-(t/CR))
.75 = (1-e^-(t/CR))
.75 -1 = -e^-(t/CR)
-0.25 = -e^-(t/CR) --->(cancelling out the negative sign)
e^-(t/CR) = 0.25
in order to remove the exponent, take logs on both sides
-t/CR = ln (0.25)
t/CR = -ln(0.25)
t = -CR x ln (0.25)
t = -(17.7 x 10^-12 x 975) x (-1.38629)
t = 23.9 x
t = 23.9ns
Thus, it took 23.9ns for the potential difference between the deflection plates to reach 75 volts