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:
distance= velocity ×time
distance= 62×10
distance=620m
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Gamma rays are the highest energy EM radiation and typically have energies greater than 100 keV, frequencies greater than 1019 Hz, and wavelengths less than 10 picometers.
Answer:
<h2>
d₂ = 3d</h2><h2>
The diameter of the second wire is 3 times that of the initial wire.</h2>
Explanation:
Using the formula for calculating the resistivity of an object to find the diameter.
Resistivity P = RA/L
R is the resistance of the material
A is the cross sectional area
L is the length of the material
Since A = πd²/4
P = R( πd²/4)/L
P = Rπd²/4L ... 1
If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;
P₂ = (R/9)A₂/L₂
P₂ = (R/9)(πd₂²/4)/L₂
P₂ = (Rπd₂²/36)/L₂
P₂ = (Rπd₂²)/36L₂
Since the length and resistivity are the same;
P = P₂ and L =L₂
Equating 1 and 2;
Rπd²/4L = (Rπd₂²)/36L₂
Rπd²/4L = (Rπd₂²)/36L
d² = d₂²/9
d₂² = 9d²
Taking the square root of both sides;
√d₂² = √9d²
d₂ = 3d
Therefore the diameter of the second wire is 3 times that of the initial wire
