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Sagot :
f. three times as much
Using the formula for kinetic energy:
KE = 1/2(mv^2)
Object 1: rest to 15 m/s
KE = 1/2(15^2)
KE = 112.5 J
Object 2: 15 m/s to 30 m/s
KE = 1/2(30^2)
KE = 450 J
450 J - 112.5 J = 337.5 J
337.5J/112.5 J = 3
Therefore the amount of energy required to accelerated the same car from 15 m/s to 30 m/s is 3 times as much as that required to accelerate the car from rest to 15 m/s.
d. four times as much, the amount of energy required to accelerate the car from 15 m/s to 30 m/s is four times as much as the amount of energy required to accelerate the same car from rest to 15 m/s
Let's evaluate the comparison of the energy required to accelerate the car from rest to 15 m/s to the energy required to accelerate it from 15 m/s to 30 m/s.
The kinetic energy [tex](\(KE\))[/tex] of an object is given by:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
where [tex]\(m\)[/tex] is the mass of the object and [tex]\(v\)[/tex] is its velocity.
For the first scenario (from rest to 15 m/s), the kinetic energy [tex](\(KE_1\))[/tex] is:
[tex]\[ KE_1 = \frac{1}{2} m (15)^2 \][/tex]
For the second scenario (from 15 m/s to 30 m/s), the kinetic energy [tex](\(KE_2\))[/tex] is:
[tex]\[ KE_2 = \frac{1}{2} m (30)^2 \][/tex]
Now, let's compare [tex]\(KE_2\) to \(KE_1\):[/tex]
[tex]\[ \frac{KE_2}{KE_1} = \frac{\frac{1}{2} m (30)^2}{\frac{1}{2} m (15)^2} \][/tex]
[tex]\[ \frac{KE_2}{KE_1} = \frac{900}{225} \][/tex]
[tex]\[ \frac{KE_2}{KE_1} = 4[/tex].
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