The strength of the fireman in vertical direction will be given by F = m * g. Then, the work done will be given by definition by W = F * d. Substituting the expression of the Force in that of the work, we have that the work will be W = m * g * d. Substituting the given values and assuming that g = 10m / s ^ 2, we have a total work of W = (73) * (10) * (9) = 6570 J
Answer:
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Explanation:
Answer:

Explanation:
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In this case, since the acceleration in terms of position is defined as its second derivative:

The purpose here is derive x(t) twice as follows:

Thus, the acceleration turns out 4.8 meters per squared seconds.
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Answer:
d = 0.38 m
Explanation:
As we know that the person due to the airbag action, comes to a complete stop, in 36 msec or less, and during this time, is decelerated at a constant rate of 60 g, we can find the initial velocity (when airbag starts to work), as follows:
vf = v₀ -a*t
If vf = 0, we can solve for v₀:
v₀ = a*t = 60*9.8 m/s²*36*10⁻³s = 21.2 m/s
With the values of v₀, a and t, we can find Δx, applying any kinematic equation that relates all of some of these parameters with the displacement.
Just for simplicity, we can use the following equation:

where vf=0, v₀ =21.2 m/s and a= -588 m/s².
Solving for d:

⇒ d = 0.38 m
Answer:
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Explanation: