Answer:
1.144 A
Explanation:
given that;
the length of the wire = 2.0 mm
the diameter of the wire = 1.0 mm
the variable resistivity R = ![\rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]](https://tex.z-dn.net/?f=%5Crho%20%28x%29%20%3D%282.5%2A10%5E%7B-6%7D%29%5B1%2B%28%5Cfrac%7Bx%7D%7B1.0%20%5C%20m%7D%29%5E2%5D)
Voltage of the battery = 17.0 v
Now; the resistivity of the variable (dR) can be expressed as = 
![dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}](https://tex.z-dn.net/?f=dR%20%3D%20%5Cfrac%7B%282.5%2A10%5E%7B-6%7D%29%5B1%2B%28%5Cfrac%7Bx%7D%7B1.0%7D%29%5E2%5D%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%2810%5E%7B-3%7D%29%5E2%7D)
Taking the integral of both sides;we have:
![\int\limits^R_0 dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5ER_0%20%20dR%20%3D%20%5Cint%5Climits%5E2_0%203.185%20%5C%20%5B1%2Bx%5E2%5D%20%5C%20dx)
![R = 3.185 [x + \frac {x^3}{3}}]^2__0](https://tex.z-dn.net/?f=R%20%3D%203.185%20%5Bx%20%2B%20%5Cfrac%20%7Bx%5E3%7D%7B3%7D%7D%5D%5E2__0)
![R = 3.185 [2 + \frac {2^3}{3}}]](https://tex.z-dn.net/?f=R%20%3D%203.185%20%5B2%20%2B%20%5Cfrac%20%7B2%5E3%7D%7B3%7D%7D%5D)
R = 14.863 Ω
Since V = IR
I = 1.144 A
∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A
Answer:
D-it's never acceptable to ignore rules and etiquette
Explanation:
Answer:
-0. 75m/s^2
Explanation:
use formula of acceleration
It uses the corkscrew to anchor it to the cork and a lever to pull the cork out
Hope this helps buddy:D
Answer:
Alpha radiation is the name for the emission of an alpha particle in fact an helium nuclei, beta radiation is the emission of electrons or positrons , and gamma radiation is the term used for the emission of energetic photons.
Explanation:
