Answer:
a) ![y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2} =640 ft+(65ft/s)(2.018s)+(\frac{1}{2})(-32.2ft/s^{2} )(2.018s)^{2}](https://tex.z-dn.net/?f=y%3Dy_%7B0%7D%2Bv_%7Boy%7D%20t%2B%5Cfrac%7B1%7D%7B2%7D%20gt%5E%7B2%7D%20%20%3D640%20ft%2B%2865ft%2Fs%29%282.018s%29%2B%28%5Cfrac%7B1%7D%7B2%7D%29%28-32.2ft%2Fs%5E%7B2%7D%20%29%282.018s%29%5E%7B2%7D)
![y=705.6ft](https://tex.z-dn.net/?f=y%3D705.6ft)
b) ![t=8.63 s](https://tex.z-dn.net/?f=t%3D8.63%20s)
Explanation:
We start the exercise knowing that a ball is thrown up with an initial velocity of 65 ft/s with an initial height of 680 ft.
To calculate the maximum heigh, we know that at the top of the motion the ball stop going up and start going down because of the <em>gravity</em>
First of all, we need to calculate the time that takes the ball to reach the maximum point.
a) ![v_{y}=v_{oy}+gt](https://tex.z-dn.net/?f=v_%7By%7D%3Dv_%7Boy%7D%2Bgt)
![t=\frac{-v_{oy} }{g}=\frac{-65ft/s}{-32.2ft/s^{2} } =2.018s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-v_%7Boy%7D%20%7D%7Bg%7D%3D%5Cfrac%7B-65ft%2Fs%7D%7B-32.2ft%2Fs%5E%7B2%7D%20%7D%20%3D2.018s)
Knowing that time, <u><em>we can calculate the height to which the ball rises:</em></u>
![y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2} =640 ft+(65ft/s)(2.018s)+(\frac{1}{2})(-32.2ft/s^{2} )(2.018s)^{2}](https://tex.z-dn.net/?f=y%3Dy_%7B0%7D%2Bv_%7Boy%7D%20t%2B%5Cfrac%7B1%7D%7B2%7D%20gt%5E%7B2%7D%20%20%3D640%20ft%2B%2865ft%2Fs%29%282.018s%29%2B%28%5Cfrac%7B1%7D%7B2%7D%29%28-32.2ft%2Fs%5E%7B2%7D%20%29%282.018s%29%5E%7B2%7D)
![y=705.6ft](https://tex.z-dn.net/?f=y%3D705.6ft)
b) Now, to know the time that the ball reach the bottom of the cliff, we know that the final height is y=0ft
![y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2}](https://tex.z-dn.net/?f=y%3Dy_%7B0%7D%2Bv_%7Boy%7D%20t%2B%5Cfrac%7B1%7D%7B2%7D%20gt%5E%7B2%7D)
![0=640ft+(65ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}](https://tex.z-dn.net/?f=0%3D640ft%2B%2865ft%2Fs%29t-%5Cfrac%7B1%7D%7B2%7D%2832.2ft%2Fs%5E%7B2%7D%29t%5E%7B2%7D)
This is a classic quadratic equation, that can be solve using the quadratic formula
![t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-b%C2%B1%5Csqrt%7Bb%5E%7B2%7D-4ac%20%7D%20%7D%7B2a%7D)
a=-16.1
b=65
c=640
Solving for t, we have that
or ![t=8.6387s](https://tex.z-dn.net/?f=t%3D8.6387s)
Since the time can not be negative:
![t=8.6387s](https://tex.z-dn.net/?f=t%3D8.6387s)
Answer:
You can determine which charge has the solid by rubbing plastic and fur, and bringing near the plastic and the solid to determine if they are repeled or attracted. If they are repeled, the solid is charged negatively and if they are attracted, the solid is charged positively.
Explanation:
In order to determine the charge of the solid (negative or positive) using a piece of plastic and fur, you have to rubb the plastic with the fur (this is charging by friction)
Negative charges are transferred from the fur to the piece of plastic, therefore the plastic is charged negatively.
Then, you have to bring near the solid and the plastic. If them are repeled, it means that the solid is charged negatively (Because similar charges are repeled)
If them are attracted, it means that the solid is charged positively (Because opposite charges are attracted)
Answer:
M V R = constant angular momentum is constant because no forces act in the direction of V
Since M (mass) = constant
V R = constant
The force is directed along the gravitational force vector (towards the center of rotation)
I halved the speed of light in vacuum to get the speed of light in the material, since the speed of light in vacuum is twice the speed of light in material.
The index of refraction is essentially the speed of light through a vacuum divided by the speed of light through given medium.
I attached some notes on refraction too, hopefully this helps :D
To solve this problem it is necessary to apply the concepts related to heat exchange and Entropy.
The temperature and mass remain constant, therefore the entropy values will be the only ones to change.
Of the three elements given their entropy values are given by
![\Delta S_{iron}=0.45J/gmC](https://tex.z-dn.net/?f=%5CDelta%20S_%7Biron%7D%3D0.45J%2FgmC)
![\Delta S_{glass}=0.8J/gmC](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bglass%7D%3D0.8J%2FgmC)
![\Delta S_{water}= 4.186J/gmC](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bwater%7D%3D%204.186J%2FgmC)
Part A) From thermodynamic theory we know that temperature is inversely proportional to entropy
Energy remains constant
![\Delta t \propto \frac{1}{S}](https://tex.z-dn.net/?f=%5CDelta%20t%20%5Cpropto%20%5Cfrac%7B1%7D%7BS%7D)
Therefore the order would be
Lowest Temperature= Water
Medium Temperature= Glass
Highest Temperature=Iron
Part B) In the case of Energy the opposite happens because it is proportional to the entropy, then
Temperature is constant
![E \propto S](https://tex.z-dn.net/?f=E%20%5Cpropto%20S)
Lowest Energy = Iron
Medium Energy = Glass
Highest Energy = Water