Explanation:
that was very good question thanks for asking.
Oil is a lubricant which helps increase friction. when u spread sand on a place where oil is spilled the sand reduces friction between road and vehicle tire. the vehicle wont skid.
D, spinal cord.
Please note that I am not 100% sure... Just taking an educated guess.
Answer:
The question is incomplete as some details are missing, here is the details ;
In this final state
a) spheres #1 and #2 both carry negative charge.
b) sphere #1 carries negative charge and #2 carries positive charge.
c) spheres #1 and #2 are still uncharged.
d) sphere #1 carries positive charge and #2 carries negative charge.
e) spheres #1 and #2 both carry positive charge.
Explanation:
From the concept of electrostatics, if a positively charged sphere is brought close to #2, there will be attraction of the opposite charges(-ve) towards it.
Now connecting a copper wire between #1 and #2, opposite charges will flow from #1 towards #2. disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charge and from the laws of attraction ; Like charges repel and unlike charges attract. the correct option is d.
A) 140 degrees
First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is
T = 32 s
So the angular velocity is
![\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D%3D%5Cfrac%7B2%5Cpi%7D%7B32%20s%7D%3D0.20%20rad%2Fs)
Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:
![\theta= \omega t](https://tex.z-dn.net/?f=%5Ctheta%3D%20%5Comega%20t)
and substituting t = 75 seconds, we find
![\theta= (0.20 rad/s)(75 s)=15 rad](https://tex.z-dn.net/?f=%5Ctheta%3D%20%280.20%20rad%2Fs%29%2875%20s%29%3D15%20rad)
In degrees, it is
![15 rad: x = 2\pi rad : 360^{\circ}\\ x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}](https://tex.z-dn.net/?f=15%20rad%3A%20x%20%3D%202%5Cpi%20rad%20%3A%20360%5E%7B%5Ccirc%7D%5C%5C%0A%3C%2Fp%3E%3Cp%3Ex%3D%5Cfrac%7B%2815%20rad%29%28360%5E%7B%5Ccirc%7D%29%7D%7B2%5Cpi%7D%3D860%5E%7B%5Ccirc%7D%20%3D%20140%5E%7B%5Ccirc%7D)
So, the new position is 140 degrees from the initial position at the top.
B) 2.7 m/s
The tangential speed, v, of a point at the egde of the wheel is given by
![v=\omega r](https://tex.z-dn.net/?f=v%3D%5Comega%20r)
where we have
![\omega=0.20 rad/s](https://tex.z-dn.net/?f=%5Comega%3D0.20%20rad%2Fs)
r = d/2 = (27 m)/2=13.5 m is the radius of the wheel
Substituting into the equation, we find
![v=(0.20 rad/s)(13.5 m)=2.7 m/s](https://tex.z-dn.net/?f=v%3D%280.20%20rad%2Fs%29%2813.5%20m%29%3D2.7%20m%2Fs)