So Hooke's law says that that law is proportional to how much I stretch the spring. Alright. So f=kx<span>. x is the length of the spring now minus its length when it's relaxed and nobody's pulling on it. k is a constant called the spring constant.</span>
Answer:
Explanation:
We shall solve this question with the help of Ampere's circuital law.
Ampere's ,law
∫ B dl = μ₀ I , B is magnetic field at distance x from the axis within wire
we shall find magnetic field at distance x . current enclosed in the area of circle of radius x
= I x π x² / π R²
= I x² / R²
B x 2π x = μ₀ x current enclosed
B x 2π x = μ₀ x I x² / R²
B = μ₀ I x / 2π R²
Maximum magnetic B₀ field will be when x = R
B₀ = μ₀I / 2π R
Given
B = B₀ / 3
μ₀ I x / 2π R² = μ₀I / 2π R x 3
x = R / 3
b ) The largest value of magnetic field is on the surface of wire
B₀ = μ₀I / 2π R
At distance x outside , let magnetic field be B
Applying Ampere's circuital law
∫ B dl = μ₀ I
B x 2π x = μ₀ I
B = μ₀ I / 2π x
Given B = B₀ / 3
μ₀ I / 2π x = μ₀I / 2π R x 3
x = 3R .
Answer:
d_{b} = 2 d_{a}
Explanation:
The electrical resistance for a cylindrical wire is described by the expression
R = ρ L / A
The area of a circle is
A = π r²
r = d / 2
A = π d²/4
We substitute
R = ρ L 4 /π d²
Let's apply this expression to our case, they indicate that the resistance of wire A is 4 times the resistance of wire B
= 4 R_{b}
We substitute
ρ 4/π 
² = 4 (ρ 4/π d_{b}²)
1 / d_{a}² = 4 / d_{b}²
d_{a} = d_{b} / 2
Answer:
I only know answer A and it's 2825.28 N/m, with rounding it's 2825.5
Explanation:
Use the m*g*h=1/2*k*x^2 equation
96*9.81*60=1/2*k*2^2
5650.56=2k
5650.56/2=2825.28N/m
B, do earthworms prefer bright light or darkness!
