Yes because if they are further away it makes it hard for them to attract each other
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
Answer:
The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg
Explanation:
Hi there!
Due to conservation of energy, the potential energy (PE) of the mass at a height of 3.32 m will be transformed into elastic potential energy (EPE) when it falls on the mattress:
PE = EPE
m · g · h = 1/2 k · x²
Where:
m = mass.
g = acceleration due to gravity.
h = height.
k = spring constant.
x = compression distance
The maximum compression distance is 0.1289 m, then, the maximum elastic potential energy will be the following:
EPE =1/2 k · x²
EPE = 1/2 · 65144 N/m · (0.1289 m)² = 541.2 J
Then, using the equation of gravitational potential energy:
PE = m · g · h = 541.2 J
m = 541.2 J/ g · h
m = 541.2 kg · m²/s² / (9.8 m/s² · 3.32 m)
m = 16.6 kg
The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg.
Answer:
Option (e)
Explanation:
A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,
Energy density = 100 J/m
Let Q be the charge on the plates.
Energy density = 1/2 x ε0 x E^2
100 = 0.5 x 8.854 x 10^-12 x E^2
E = 4.75 x 10^6 V/m
V = E x d
V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V
C = ε0 A / d
C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F
Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C
Q = 190 nC