Answer:
The speed of the car when load is dropped in it is 10.90 m/s.
Explanation:
Given that,
Mass of the railroad car, m₁ = 11600kg
Mass of the load, m₂ = 5420kg
It can be assumed as the speed of the car, u₁ = 16 m/s
Initially, it is at rest, u₂ = 0
The additional load is dropped onto the car.
Let v is the speed of the car. It can be calculated using the conservation of momentum as :
So, the speed of the car when load is dropped in it is 10.90 m/s.
Positive will react better together. But opposites will try to get as far away as possible.
Explanation:
It is given that,
Current in wire 1, I₁ = 10 A
Current in wire 2, I₂ = 20 A
Distance between wires, d = 10 cm = 0.1 m
Force per unit length is given by :
So, the magnetic force acting per unit length of the wires 
. Since, the current is in same direction. So, the force is attractive in nature.
Answer:
6.54 × 10⁻⁵ Pa-s
Explanation:
Since the shear force, F = μAu/y where μ = viscosity of fluid between plates, A = area of plates, u = velocity of fluid = 0.6 m/s and y = separation of plates = 0.02 mm = 2 × 10⁻⁵ m
Since F = μAu/y
F/A = μu/y where F/A = force per unit area
Since we are given force per unit area, F/A = 1.962 N per unit area = 1.962 N/m²
So, μ = F/A ÷ u/y
substituting the values of the variables into the equation, we have
μ = F/A ÷ u/y
μ = 1.962 N/m² ÷ 0.6 m/s/2 × 10⁻⁵ m
μ = 1.962 N/m² ÷ 0.3 × 10⁵ /s
μ = 6.54 × 10⁻⁵ Ns/m²
μ = 6.54 × 10⁻⁵ Pa-s
