There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:
![v=15 m/s](https://tex.z-dn.net/?f=v%3D15%20m%2Fs)
- radius of the hill:
![r=100 m](https://tex.z-dn.net/?f=r%3D100%20m)
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
![W=mg](https://tex.z-dn.net/?f=W%3Dmg)
(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force,
![m \frac{v^2}{r}](https://tex.z-dn.net/?f=m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
, so we can write:
![mg-N=m \frac{v^2}{r}](https://tex.z-dn.net/?f=mg-N%3Dm%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
(1)
By rearranging the equation and substituting the numbers, we find N:
![N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N](https://tex.z-dn.net/?f=N%3Dmg-m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%3D%28975%20kg%29%289.81%20m%2Fs%5E2%29-%28975%20kg%29%20%5Cfrac%7B%2815%20m%2Fs%29%5E2%7D%7B100%20m%7D%3D7371%20N%20%20)
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
![N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N](https://tex.z-dn.net/?f=N%3Dmg-m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%3D%2862%20kg%29%289.81%20m%2Fs%5E2%29-%2862%20kg%29%20%5Cfrac%7B%2815%20m%2Fs%29%5E2%7D%7B100%20m%7D%3D469%20N%20)
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
![mg=m \frac{v^2}{r}](https://tex.z-dn.net/?f=mg%3Dm%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
from which we find
Answer:
E.Potential energy is transformed into kinetic and thermal energy.
Explanation:
At the top of the hill, the roller coaster car has gravitational potential energy, which is given by:
![U=mgh](https://tex.z-dn.net/?f=U%3Dmgh)
where m is the mass of the car, g is the acceleration due to gravity and h is the height of the car from the ground.
As the car moves down, its height h decreases, while its speed v increases: therefore, potential energy is converted into kinetic energy:
![K=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where v is the speed of the car.
However, we must also take into account the force of friction between the wheels of the car and the roller coaster track. This force slows down the car, and so some of the car's energy is actually "lost" and converted into thermal energy (which is due to the friction between the car wheels and the track).
Easy, it is called:
The intensity.
I thinks or know it is A or B cuz magnets and metal attract
<span>(A)
s{ t } = (¼)t² + 2t + 1 ... differentiate to get velocity versus time
v{ t } = (½)t + 2 ... linearly increases with time
(1)
Vavɢ{ a→b } = [ v{ b } + v{ a } ] ⁄ 2
Vavɢ{ 1→3 } = [ v{ 3 } + v{1 } ] ⁄ 2
Vavɢ{ 1→3 } = [ (½)(3) + 2 + (½)(1) + 2 ] ⁄ 2
Vavɢ{ 1→3 } = 3 m/sec
(II) and (III) are done the same way.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~...
(B)
v{ t } = (½)t + 2
v{1 } = (½)(1) + 2
v{1 } = 2.5 m/sec</span>