A chain position over a certain amount of time
Explanation:
Given that,
Initial speed of the car, u = 88 km/h = 24.44 m/s
Reaction time, t = 2 s
Distance covered during this time, 
(a) Acceleration, 
We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :


s = 74.66 meters
s = 74.66 + 48.88 = 123.54 meters
(b) Acceleration, 


s = 37.33 meters
s = 37.33 + 48.88 = 86.21 meters
Hence, this is the required solution.
First thing to note is that we are on week 0, when they start saving:
The equation for Alex is:
y=20x + 125
Know that we are on week 0, and x is our weeks, that makes our equation:
y=20(0) + 125
Which means it's really:
y=125
That means that the first one is E. 125
For the second, it shows our week 0, which is at 100.
That means that the second one is I. 100
Hope this helps!
Answer: 
Explanation:
Given
Cross-sectional area 
Dielectric constant 
Dielectric strength 
Distance between capacitors 
Maximum charge that can be stored before dielectric breakdown is given by
![\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC](https://tex.z-dn.net/?f=%5CRightarrow%20Q%3DCV%5C%5C%5C%5C%5CRightarrow%20Q%3D%5Cdfrac%7Bk%5Cepsilon_oA%7D%7Bd%7D%5Ccdot%20%28Ed%29%5Cquad%5Cquad%20%5BV%3DE%5Ccdot%20d%5D%5C%5C%5C%5C%5CRightarrow%20Q%3Dk%5Cepsilon_oAE%5C%5C%5C%5C%5CRightarrow%20Q%3D4%5Ctimes%208.85%5Ctimes%2010%5E%7B-12%7D%5Ctimes%200.4%5Ctimes%2010%5E%7B-4%7D%5Ctimes%202%5Ctimes%2010%5E8%5C%5C%5C%5C%5CRightarrow%20Q%3D28.32%5Ctimes%2010%5E%7B-8%7D%5C%5C%5C%5C%5CRightarrow%20Q%3D283.2%5Ctimes%2010%5E%7B-9%7D%5C%20nC)