What is the power of 10 when 0.00503 is written in scientific notation
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m = mass = 5 kg
= initial velocity = 100 m/s
= final velocity = ?
I = impulse = 30 Ns
Using the impulse-change in momentum equation
I = m( -
)
30 = 5 ( - 100)
= 106 m/s
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)