Answer:
The magnitude of the force between the two parallel wires is 0.0111 N.
Explanation:
Given;
length of the two parallel wires, L = 42 m
distance between the two wires, r = 0.03 m
current in both wires, I₁, I₂ = 6.3 A
Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.
Answer:
d = 52 μm
Explanation:
given,
wavelength of the light source (λ)= 550 nm
distance to form interference pattern(D) = 1.5 m
y = 1.6 cm = 0.016 m
width of the slits = ?
now, using displacement formula
for the first maxima, m = 1
d = 5.2 x 10⁻⁶ m
d = 52 μm
hence, the width of her slits is equal to d = 52 μm
Answer:
Gamma rays
Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies.
Explanation:
The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k
Let r be the vector perpendicular to A and B,
r = A * B
A = 3i + 6j - 2k
B = 4i - j + 3k
a1 = 3
a2 = 6
a3 = - 2
b1 = 4
b2 = - 1
b3 = 3
a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k
a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k
a * b = 16 i - 17 j - 27 k
The perpendicular vector, r = 16 i - 17 j - 27 k
Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k
To know more about perpendicular vectors
brainly.com/question/14384780
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