The velocity of B after elastic collision is 3.45m/s
This type of collision is an elastic collision and we can use a formula to solve this problem.
<h3>Elastic Collision</h3>

The data given are;
- m1 = 281kg
- u1 = 2.82m/s
- m2 = 209kg
- u2 = -1.72m/s
- v1 = ?
Let's substitute the values into the equation.

From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.
Learn more about elastic collision here;
brainly.com/question/7694106
Answer:
Cuba, Jamaica, Dominican Republic, Haiti, Turks & Caicos, and The Bahamas
Explanation:
Answer:
The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.
Explanation:
Given;
distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m
current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively
The magnitude of the magnetic field halfway between the wires can be calculated as;

where;
B is magnitude of the magnetic field halfway between the wires
I₁ is current in the first wire
I₂ is current the second wire
μ₀ is permeability of free space
r is distance half way between the wires

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.