Answer:
1.F is the electrostatic force between charges (in Newtons),
2.q₁ is the magnitude of the first charge (in Coulombs),
3.q₂ is the magnitude of the second charge (in Coulombs),
4.r is the shortest distance between the charges (in m),
5.ke is the Coulomb's constant. It is equal to 8.98755 × 10⁹ N·m²/C² .
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:
8400m
Explanation:
The engine that falls off would have the same constant horizontal velocity as the airplane's when if falls off if we ignore air resistance. So it would have a horizontal velocity of 280m/s for 30seconds before it hits the ground.
Therefor the horizontal distance the engine travels during its fall is
280 * 30 = 8400m
Answer:
F = 2 * 30 / 5 = 12 N to stop forward motion
F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees
(12^2 + 16^2)^1.2 = 20 N average force applied