Answer:
the velocity of the boats after the collision is 4.36 m/s.
Explanation:
Given;
mass of fish, m₁ = 800 kg
mass of boat, m₂ = 1400 kg
initial velocity of the fish, u₁ = 12 m/s
initial velocity of the boat, u₂ = 0
let the final velocity of the fish-boat after collision = v
Apply the principle of conservation of linear momentum for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
800 x 12 + 1400 x 0 = v(800 + 1400)
9600 = 2200v
v = 9600/2200
v = 4.36 m/s
Therefore, the velocity of the boats after the collision is 4.36 m/s.
Answer:
m= 10 kg a = 52 m / s²
Explanation:
For this problem we must use Newton's second law, let's apply it to each axis
X axis
F - fr = ma
The equation for the force of friction is
-fr = miu N
Axis y
N- W = 0
N = mg
Let's replace and calculate laceration
F - miu (mg) = ma
a = F / m - mi g
a = 527.018 / m - 0.17 9.8
We must know the mass of the body suppose m = 10 kg
a = 527.018 / 10 - 1,666
a = 52 m / s²
The acceleration is -9.8m/s^2. The initial velocity is -8m/s. The initial position is 30m. This describes a position function of
-(9.8/2)t^2-8t+30=0
Solve the quadratic equation for t to get t=1.789s
The Net Force would be 2 N to the left.
21 N is being used to push the box to the right and 23 N is used to push it left. There is a stronger force pushing the box towards the left. The different in the two numbers would give you the net force acting on the box and the direction of the arrow with the greatest force will tell you the direction.
