Answer:
The average velocity is
and 
respectively.
Explanation:
Let's start writing the vertical position equation :
Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt = 
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :
For the position variation we use the vertical position equation :
Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is
For the second time interval :
t1 = 4 s → t2 = 9 s
Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :
Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then
The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is
Complete Question
A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?
Answer:
The values is 
Explanation:
From the question we are told that
The diameter is 
The charge is 
The distance from the center is 
Generally the radius is mathematically represented as
=> 
=> 
Generally electric field is mathematically represented as
![E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7BQ%7D%7B%202%5Cepsilon_o%20%7D%20%5B1%20-%20%5Cfrac%7Bk%7D%7B%5Csqrt%7Br%5E2%20%2B%20%20k%5E2%20%7D%20%7D%20%5D)
substituting values
![E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B4.4%20%2A10%5E%7B-9%7D%7D%7B%202%2A%20%288.85%2A10%5E%7B-12%7D%29%20%7D%20%5B1%20-%20%5Cfrac%7B%281.00%20%2A10%5E%7B-4%7D%29%7D%7B%5Csqrt%7B%280.06%29%5E2%20%2B%20%20%281.0%2A10%5E%7B-4%7D%29%5E2%20%7D%20%7D%20%5D)
Are there any options or is it not multiple choice.
Answer:
1.3 x 10^(-2) atm/s
Explanation:
It follows the stoichiometry. For every mole of O3 that disappears, 1.5 moles (that is, 3/2) of O2 appears:
1.5 * 0.009 atm/s = 0.0135 atm/sec; the answer is 1.3 x 10^(-2) atm/s
The rate at which the height is changing is ( 5 / x ) m / hr
We know that,
Area of an equilateral triangle A = 
/ 4
h = x / 2
Where,
x = Side
h = Height
Given that,
dA / dt = 5 
/ hr
h = x / 2
Differentiate both sides with respect to t
dh / dt = ( / 2 ) ( dx / dt )
dx / dt = ( 2 / ) ( dh / dt )
A = / 4
Differentiate both sides with respect to t
dA / dt = ( / 4 ) ( 2x ) ( dx / dt )
5 = ( / 4 ) ( 2x ) ( 2 / ) ( dh / dt )
dh / dt = ( 5 / x ) m / hr
Rate of change of height is defined as the rate at which height of an object changes with respect to time. It is represented as dh / dt
Therefore, the rate at which the height is changing is ( 5 / x ) m / hr
To know more about Rate of change of height
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