Answer: Depends
Explanation:
Depends on how much the diver weighs.
<span>Data:
mass =
110-g bullet
d = 0.636 m
Force =
13500 + 11000x - 25750x^2, newtons.
a) Work, W
W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =
W = 13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636 =
W = 8602.6 joule
b) x= 1.02 m
</span><span><span>W = 13500x + 5500x^2 - 8583.33 x^3 ] from</span> 0 to 1.02
W = 10383.5
c) %
[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.
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A perfectly elastic<span> collision is defined as one in which there is no loss of </span>kinetic energy<span> in the collision. Therefore, we just add the kinetic energies of each system. We calculate as follows:
KE = 0.5(</span>1.0 × 10^3)(12.5 )^2 + 0.5(1.0 × 10^3)(12.5 )^2
KE = 156250 J = 1.6 x 10^5 J -------> OPTION A
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