...the potential energy that you build while going up the hill on the roller coaster could be let go as kinetic energy -- the energy of motion that takes you down the hill of the roller coaster.
Answer:
Cam Newton (currently but might change because he has been allowed to trade)
Will Grier
Kyle Allen
Explanation:
Answer: 29.17m/s^2
Explanation:
Given the following :
Velocity = 525 m/s
Time = 18 seconds
Acceleration = change in Velocity with time
Using the motion equation:
v = u + at
Where v = final Velocity
u = Initial Velocity and t = time
Plugging our values
525 = 0 + a × 18
525 = 18(a)
a = 525 / 18
a = 29.166666
a = 29.17 m/s^2
If the compressor removes 500 j if the 700 the 200 j left would have Ben conducted through the refrigerator/200 j released into the room
Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis
![F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}](https://tex.z-dn.net/?f=F_N%3Dk_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D)
Explanation:
Let the force on +Q charge y-axis due to +2Q charge be 
and force on +Q charge y axis due to -Q charge on x-axis be 
.
Distance between the +2Q charge and +Q charge = d units
Distance between the -Q charge and +Q charge = 
units
= Coulomb constant
Net force on +Q charge on y-axis is:
![|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|](https://tex.z-dn.net/?f=%7CF_N%7C%3D%7Ck_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D%7C)
The net froce on the +Q charge on y-axis is
