Answer:
- KEi = 2.256×10^5 J
- KEf = 9.023×10^5 J
- 4 times as much work
Step-by-step explanation:
The kinetic energy for a given mass and velocity is ...
KE = (1/2)mv^2 . . . . . m is mass
At its initial speed, the kinetic energy of the car is ...
KEi = (1/2)(810 kg)(23.6 m/s)^2 ≈ 2.256×10^5 J . . . . . m is meters
At its final speed, the kinetic energy of the car is ...
KEf = (1/2)(810 kg)(47.2 m/s)^2 ≈ 9.023×10^5 J
The ratio of final to initial kinetic energy is ...
(9.023×10^5)/(2.256×10^5) = 4
4 times as much work must be done to stop the car.
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You know this without computing the kinetic energy. KE is proportional to the square of speed, so when the speed doubles, the KE is multiplied by 2^2 = 4.
B is the right choice.
You move the point to two spaces right, making 19.28 (percent).
It's difficult to make out, but I think the task is to expand
Write the number in polar form first:
By DeMoivre's theorem, you have
and converting back to Cartesian form, this number is equivalent to
Instead of distributing 2, we can actually divide both sides by 2:
x - 1 = 2x - 11
Isolate x:
x = 10
