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The impulse of a force is due to the change in the motion of an object
A. The persons speed after impact is approximately 59.38 mi/h
B. The expected speed is <u>29.89 mi/h</u> which is less than the findings
Reason:
Known parameters are;
The speed of the bus, v = 30 mi/h
The force with which the person was hit, F = 58,000 lbs
Mass of the bus, M = 40,000 lbs
Mass of the person, m = 150 lbs
Duration of the impact, Δt = 0.007 seconds
A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;
The impulse of the force = F × Δt = m × Δv
For the person, we get;
58,000 lbf ≈ 1866094.816 lb·ft./s²
58,000 lbf × 0.007 s = 150 lbs × Δv
1,866,094.816 lb·ft./s²
Δv = v₂ - v₁
The initial speed of the person at the instant, can be as v₁ = 0
The final speed, v₂ = Δv - v₁
∴ v₂ ≈ 87.084 ft./s - 0 = 87.084 ft./s
≈ <u>87.084 ft./s</u>
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The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>
B. Where the momentum is conserved, we have;
m₁·v₁ + m₂v₂ = (m₁ + m₂)·v
The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>
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Answer:
A. h = 2.15 m
B. Pb' = 122 KPa
Explanation:
The computation is shown below:
a) Let us assume the depth be h
As we know that
After solving this,
h = 2.15 m
Therefore the depth of the fluid is 2.15 m
b)
Given that
height of the extra fluid is
h' = 0.355 m
Now let us assume the pressure at the bottom is Pb'
so, the equation would be
Pb' = 122 KPa