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.
_____
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.
The answer is : B. (X-10)(x-4)
5*y+3? is that the answer u was looking for?
Answer:
t0 = π/4
Step-by-step explanation:
The velocity v(t) = r'(t) = <-2sint,-2cost>
Since <1,-1>⊥v, <1,-1>•v=0. So,
(1)(-2sint) + (-1)(-2cost) = 0
cost - sint = 0
