Answer:
Explanation:
Two straight wires
Have current in opposite direction
i1=i2=i=2Amps
Distance between two wires
r=5mm=0.005m
Length of one wire is ∞
Length of second wire is 0.3m
Force between the wire,
The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by
F/l = μoi1i2/2πr
F/l=μoi²/2πr
μo=4π×10^-7 H/m
The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.
F/l = μoi1i2/2πr
F/0.3=4π×10^-7×2²/2π•0.005
F/0.3=1.6×10^-4
Cross multiply
F=1.6×10^-4×0.3
F=4.8×10^-5N
Answer:
3 x 10^5 J
Explanation:
mass of substance, m = 1 g = 0.001 kg
Velocity of light, c = 3 x 10^8 m/s
According to the Einstein mass energy equivalence, the energy associated with the mass is given by
E = m c^2
E = 0.001 x 3 x 10^8
E = 3 x 10^5 J
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
