How does the angle of launch affect the kinetic energy of a rubber band?
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:
Temperature increase = 2.1 [C]
Explanation:
We need to identify the initial data of the problem.
v = velocity of the copper sphere = 40 [m/s]
Cp = heat capacity = 387 [J/kg*C]
The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be: