Suppose an automobile has 2000-joules of kinetic energy. when it moves at twice the speed, what will be its kinetic energy? what
1 answer:
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Refer to the diagram shown below.
g = 9.8 m/s², and air resistance is ignored.
For mass m₁:
The normal reaction is m₁g.
The resisting force is R₁ = μm₁g.
For mass m₂:
The normal reaction is m₂g.
The resisting force is R₂ = μm₂g.
Let a = the acceleration of the system.
Then
(m₁ + m₂)a = F - (R₁ + R₂)
(14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg)
40a = 65 - 38.416 = 26.584
a = 0.6646 m/s²
Answer: 0.665 m/s² (nearest thousandth)
g = 9.8 m/s², and air resistance is ignored.
For mass m₁:
The normal reaction is m₁g.
The resisting force is R₁ = μm₁g.
For mass m₂:
The normal reaction is m₂g.
The resisting force is R₂ = μm₂g.
Let a = the acceleration of the system.
Then
(m₁ + m₂)a = F - (R₁ + R₂)
(14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg)
40a = 65 - 38.416 = 26.584
a = 0.6646 m/s²
Answer: 0.665 m/s² (nearest thousandth)
Answer:
Explanation:
1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is
100kg/h*0.5h = 50kg
2. the momentum does not conserve because the drag force of water makes that the boat loses velocity
3. If we assume that the force of the boat before the raining is
where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts
And if we take the net force as
where we take v=1m/s because we are taking into account tha velocity just after the rain stars.
I hope this is useful for you
regards