Answer:
1- For the track B. The potential energy is the same for the two cars, but because of the slope of the track, the car B earn kinetic energy faster. The gravitation acceleration of the cars will be g•sinθ, and the angle of the track B will have a bigger value for sinθ
2- The conservation of energy applies because the roller coaster is a closed track. When a car climb the track, it earn GPE, which is given by mgh, when it get down in the track, it transform GPE in KE, which is given in 1/2mv².
3-
Position of car (m) GPE KE GPE + KE
top (30m) 60000 0 60000
bottom (0m) 0 60000 60000
halfway down (15m) 30000 30000 60000
three-quarters way down 15000 45000 60000
Answer:
(a) ![\overrightarrow{L}=885.5\widehat{k}](https://tex.z-dn.net/?f=%5Coverrightarrow%7BL%7D%3D885.5%5Cwidehat%7Bk%7D)
(b) ![\overrightarrow{L}=1046.5\widehat{k}](https://tex.z-dn.net/?f=%5Coverrightarrow%7BL%7D%3D1046.5%5Cwidehat%7Bk%7D)
Explanation:
mass, m = 2.3 kg
vx = 40 m/s
vy = 75 m/s
(a) Angular momentum is given by
![\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}](https://tex.z-dn.net/?f=%5Coverrightarrow%7BL%7D%3D%5Coverrightarrow%7Br%7D%5Ctimes%20%5Coverrightarrow%7Bp%7D)
Where, p is the linear momentum and r is the position vector about which the angular momentum is calculated.
Here, ![\overrightarrow{r}=3\widehat{i}-4\widehat{j}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Br%7D%3D3%5Cwidehat%7Bi%7D-4%5Cwidehat%7Bj%7D)
![\overrightarrow{p}=m\overrightarrow{v}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bp%7D%3Dm%5Coverrightarrow%7Bv%7D)
![\overrightarrow{p}=2.3\left ( 40\widehat{i}+75\widehat{j} \right )](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bp%7D%3D2.3%5Cleft%20%28%2040%5Cwidehat%7Bi%7D%2B75%5Cwidehat%7Bj%7D%20%5Cright%20%29)
![\overrightarrow{p}= 92\widehat{i}+172.5\widehat{j}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bp%7D%3D%2092%5Cwidehat%7Bi%7D%2B172.5%5Cwidehat%7Bj%7D)
So, the angular momentum
![\overrightarrow{L}=\left ( 3\widehat{i}-4\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )](https://tex.z-dn.net/?f=%5Coverrightarrow%7BL%7D%3D%5Cleft%20%28%203%5Cwidehat%7Bi%7D-4%5Cwidehat%7Bj%7D%20%5Cright%20%29%5Ctimes%5Cleft%20%28%2092%5Cwidehat%7Bi%7D%2B172.5%5Cwidehat%7Bj%7D%20%5Cright%20%29)
![\overrightarrow{L}=885.5\widehat{k}](https://tex.z-dn.net/?f=%5Coverrightarrow%7BL%7D%3D885.5%5Cwidehat%7Bk%7D)
(b) Here, ![\overrightarrow{r}=(3+2)\widehat{i}+(-4+2)\widehat{j}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Br%7D%3D%283%2B2%29%5Cwidehat%7Bi%7D%2B%28-4%2B2%29%5Cwidehat%7Bj%7D)
![\overrightarrow{r}=5\widehat{i}-2\widehat{j}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Br%7D%3D5%5Cwidehat%7Bi%7D-2%5Cwidehat%7Bj%7D)
![\overrightarrow{L}=\left ( 5\widehat{i}-2\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )](https://tex.z-dn.net/?f=%5Coverrightarrow%7BL%7D%3D%5Cleft%20%28%205%5Cwidehat%7Bi%7D-2%5Cwidehat%7Bj%7D%20%5Cright%20%29%5Ctimes%5Cleft%20%28%2092%5Cwidehat%7Bi%7D%2B172.5%5Cwidehat%7Bj%7D%20%5Cright%20%29)
![\overrightarrow{L}=1046.5\widehat{k}](https://tex.z-dn.net/?f=%5Coverrightarrow%7BL%7D%3D1046.5%5Cwidehat%7Bk%7D)
The molecules separate and become atoms and there magnetic forces deprecate and no longer collide
If you have a lump of solid at its melting point ... like ice at 32°F ...
you have to put a certain amount of heat into it just to change it
to water at 32°F. That amount of heat, that's used just to change
a solid lump into liquid without changing its temperature, is called
the heat of fusion for that substance.
The number is different for every substance.
For water, it takes 336 joules of heat to melt 1 gram of ice
into 1 gram of water, all at 32°F (0°C).
That's an enormous latent heat of fusion ... more than almost any
other known substance. That's why ice is such a good choice
when you need something to put in your drink to cool it down.
Ice absorbs a huge amount of heat before it melts and the drink
gets watered down.