Answer:
Resistance = 68.23 Ω
Explanation:
Let's start off by remembering the fact that when resistors are connected in series, their resistances is added up to find the total resistance.
The equation for resistance using resistivity is given below:
Resistance = Resistivity * Length / Area
where Length = x
Resistivity = 3x^5
and Area =
= 8.04 * 10^(-8) meter squared
Substituting in the value of resistivity, length and area we get:
Resistance = ![\frac{ (3x^5) * (x) }{(8.04*10^-^8)}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%283x%5E5%29%20%2A%20%28x%29%20%7D%7B%288.04%2A10%5E-%5E8%29%7D)
Resistance = ![\frac{ (3x^6) }{(8.04*10^-^8)}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%283x%5E6%29%20%7D%7B%288.04%2A10%5E-%5E8%29%7D)
Since resistance in series is added, we can simply integrate this formula over the length (x = 0 to x = 0.2) to get the total resistance.
Resistance = ![\int\ {\frac{ (3x^6) }{(8.04*10^-^8)}} \, dx](https://tex.z-dn.net/?f=%5Cint%5C%20%7B%5Cfrac%7B%20%283x%5E6%29%20%7D%7B%288.04%2A10%5E-%5E8%29%7D%7D%20%5C%2C%20dx)
Resistance = ![(5.4857*10^-^6) /(8.04*10^-^8)](https://tex.z-dn.net/?f=%285.4857%2A10%5E-%5E6%29%20%2F%288.04%2A10%5E-%5E8%29)
Resistance = 68.23 Ω
Answer:
- The formula its
![f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816](https://tex.z-dn.net/?f=f%28t%29%20%5C%20%3D%20%5C%20-%20%5C%20352%20%5C%20%5Cfrac%7B%5C%24%20%7D%7Byears%7D%20%5C%20t%20%5C%20%2B%20%5C%20%5C%24%20%5C%202816%20)
- After 5 years, the computer value its $ 1056
Explanation:
<h3>
Obtaining the formula</h3>
We wish to find a formula that
- Starts at 2816.
![f(0 \ years) \ = \ \$ \ 2816](https://tex.z-dn.net/?f=f%280%20%5C%20years%29%20%5C%20%3D%20%5C%20%5C%24%20%5C%202816)
- Reach 0 at 8 years.
![f( 8 \ years) \ = \ \$ \ 0](https://tex.z-dn.net/?f=f%28%208%20%5C%20years%29%20%5C%20%3D%20%5C%20%5C%24%20%5C%200)
- Depreciates at a constant rate. m
We can cover all this requisites with a straight-line equation. (an straigh-line its the only curve that has a constant rate of change) :
,
where m its the slope of the line and b give the place where the line intercepts the <em>y</em> axis.
So, we can use this formula with the data from our problem. For the first condition:
![f ( 0 \ years ) = m \ (0 \ years) + b = \$ \ 2816](https://tex.z-dn.net/?f=f%20%28%200%20%5C%20years%20%29%20%3D%20m%20%5C%20%280%20%5C%20years%29%20%2B%20b%20%3D%20%5C%24%20%5C%202816)
![b = \$ \ 2816](https://tex.z-dn.net/?f=%20b%20%3D%20%5C%24%20%5C%202816)
So, b = $ 2816.
Now, for the second condition:
![f ( 8 \ years ) = m \ (8 \ years) + \$ \ 2816 = \$ \ 0](https://tex.z-dn.net/?f=f%20%28%208%20%5C%20years%20%29%20%3D%20m%20%5C%20%288%20%5C%20years%29%20%2B%20%5C%24%20%5C%202816%20%3D%20%5C%24%20%5C%200)
![m \ (8 \ years) = \ - \$ \ 2816](https://tex.z-dn.net/?f=%20m%20%5C%20%288%20%5C%20years%29%20%3D%20%5C%20-%20%5C%24%20%5C%202816)
![m = \frac{\ - \$ \ 2816}{8 \ years}](https://tex.z-dn.net/?f=%20m%20%3D%20%5Cfrac%7B%5C%20-%20%5C%24%20%5C%202816%7D%7B8%20%5C%20years%7D)
![m = \frac{\ - \$ \ 2816}{8 \ years}](https://tex.z-dn.net/?f=%20m%20%3D%20%5Cfrac%7B%5C%20-%20%5C%24%20%5C%202816%7D%7B8%20%5C%20years%7D)
![m = \ - \ 352 \frac{\$ }{years}](https://tex.z-dn.net/?f=%20m%20%3D%20%5C%20-%20%5C%20352%20%5Cfrac%7B%5C%24%20%7D%7Byears%7D)
So, our formula, finally, its:
![f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816](https://tex.z-dn.net/?f=f%28t%29%20%5C%20%3D%20%5C%20-%20%5C%20352%20%5C%20%5Cfrac%7B%5C%24%20%7D%7Byears%7D%20%5C%20t%20%5C%20%2B%20%5C%20%5C%24%20%5C%202816%20)
<h3>After 5 years</h3>
Now, we just use <em>t = 5 years</em> in our formula
![f(5 \ years) \ = \ - \ 352 \ \frac{\$ }{years} \ 5 \ years \ + \ \$ \ 2816](https://tex.z-dn.net/?f=f%285%20%5C%20years%29%20%5C%20%3D%20%5C%20-%20%5C%20352%20%5C%20%5Cfrac%7B%5C%24%20%7D%7Byears%7D%20%5C%205%20%5C%20years%20%5C%20%2B%20%5C%20%5C%24%20%5C%202816%20)
![f(5 \ years) \ = \ - \$ \ 1760 + \ \$ \ 2816](https://tex.z-dn.net/?f=f%285%20%5C%20years%29%20%5C%20%3D%20%5C%20-%20%5C%24%20%5C%201760%20%2B%20%5C%20%5C%24%20%5C%202816%20)
![f(5 \ years) \ = $ \ 1056](https://tex.z-dn.net/?f=f%285%20%5C%20years%29%20%5C%20%3D%20%24%20%5C%201056%20)
Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
<h3>What is electric force?</h3>
Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.
![\rm F= K\frac{q_2q_3}{r^2}](https://tex.z-dn.net/?f=%5Crm%20F%3D%20K%5Cfrac%7Bq_2q_3%7D%7Br%5E2%7D)
![\rm F= 9\times 10^9 \times \frac{1.6 \times 10^{-13}\times 1.6\times10^{-13}}{(0.5)^2}](https://tex.z-dn.net/?f=%5Crm%20F%3D%209%5Ctimes%2010%5E9%20%5Ctimes%20%5Cfrac%7B1.6%20%5Ctimes%2010%5E%7B-13%7D%5Ctimes%201.6%5Ctimes10%5E%7B-13%7D%7D%7B%280.5%29%5E2%7D)
![\rm F= - 1. 1 \times 10^{11 }N](https://tex.z-dn.net/?f=%5Crm%20F%3D%20%20-%201.%201%20%5Ctimes%2010%5E%7B11%20%7DN)
Hence the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
To learn more about the electric force refer to the link;
brainly.com/question/1076352
Answer:
The answer is C
Explanation:
When the project finishes under a certain agreement. the project team must capture and share lessons that learn to not repeat same mistakes that made while project ongoing. Therefore, when the similar work need to be redone new project team saves time and labor force with those feedback.
The project team must secure customer feedback and approval to get their claims and progress payments.
The project team must plan a smooth transition of deliverable into ongoing operations in order to get feedback or approval from them in a certain of time. With those feedback of approvals the project team can plan their rework or payment progress.
Alternative dispute resolution is not a must work. It depends on the agreement.