![Voltage = 1.47\\L= 22.6m](https://tex.z-dn.net/?f=Voltage%20%3D%201.47%5C%5CL%3D%2022.6m)
![r=\frac{d}{2} = \frac{0.04in}{2} = 0.02in = 0.508mm = 0.508*10^{-3}m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bd%7D%7B2%7D%20%3D%20%5Cfrac%7B0.04in%7D%7B2%7D%20%3D%200.02in%20%3D%200.508mm%20%3D%200.508%2A10%5E%7B-3%7Dm)
So we can now calculate the area of the wire
![A= \pi *r^2 = \pi (0.508*10^{-3})^2 = 8.107*10^{-7}m^2](https://tex.z-dn.net/?f=A%3D%20%5Cpi%20%2Ar%5E2%20%3D%20%5Cpi%20%280.508%2A10%5E%7B-3%7D%29%5E2%20%3D%208.107%2A10%5E%7B-7%7Dm%5E2)
We need also the resistivity specific of cupper, that is:
![p_{cupper}=1,72*10^{-8}](https://tex.z-dn.net/?f=p_%7Bcupper%7D%3D1%2C72%2A10%5E%7B-8%7D)
a) We have to calculate the current that is given by,
![R=\frac{pL}{A}\\R=\frac{(1.72*10^{8})(22.6)}{(8.107*10^{-7})}\\R=0.4\Omega](https://tex.z-dn.net/?f=R%3D%5Cfrac%7BpL%7D%7BA%7D%5C%5CR%3D%5Cfrac%7B%281.72%2A10%5E%7B8%7D%29%2822.6%29%7D%7B%288.107%2A10%5E%7B-7%7D%29%7D%5C%5CR%3D0.4%5COmega)
b)![I=\frac{V}{R}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BV%7D%7BR%7D)
![I=\frac{1.47}{0.4} = 3.675A](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B1.47%7D%7B0.4%7D%20%3D%203.675A)
c) current density
![j= \frac{I}{A}\\j= \frac{3.675}{8.107*10^{-7}}\\j= 4.533*10^5 A/m^2](https://tex.z-dn.net/?f=j%3D%20%5Cfrac%7BI%7D%7BA%7D%5C%5Cj%3D%20%5Cfrac%7B3.675%7D%7B8.107%2A10%5E%7B-7%7D%7D%5C%5Cj%3D%204.533%2A10%5E5%20A%2Fm%5E2)
d) Rate of thermal energy
![\alpha = I^2*R\\\alpha = 3.675^2*0.4 = 5.4W](https://tex.z-dn.net/?f=%5Calpha%20%3D%20I%5E2%2AR%5C%5C%5Calpha%20%3D%203.675%5E2%2A0.4%20%3D%205.4W)
Answer:
![E=8*10^5\frac{V}{m}](https://tex.z-dn.net/?f=E%3D8%2A10%5E5%5Cfrac%7BV%7D%7Bm%7D)
Explanation:
The magnitude of the electric field between two parallel conducting plates is defined as:
![E=\frac{\Delta V}{d}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B%5CDelta%20V%7D%7Bd%7D)
Here
is the potential difference between the plates and d its separation.
The electric potential energy is defined as the product between the particle's charge and the potential difference:
![U=q\Delta V](https://tex.z-dn.net/?f=U%3Dq%5CDelta%20V)
Solving for
and replacing in the electric field formula:
![\Delta V=\frac{U}{q}\\E=\frac{U}{qd}](https://tex.z-dn.net/?f=%5CDelta%20V%3D%5Cfrac%7BU%7D%7Bq%7D%5C%5CE%3D%5Cfrac%7BU%7D%7Bqd%7D)
In this case we have a double charged ion, so
:
![E=\frac{32*10^3eV}{(2e)(2*10^{-2}m)}\\E=8*10^5\frac{V}{m}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B32%2A10%5E3eV%7D%7B%282e%29%282%2A10%5E%7B-2%7Dm%29%7D%5C%5CE%3D8%2A10%5E5%5Cfrac%7BV%7D%7Bm%7D)
Answer:
a = -47.93 m/s²
Explanation:
given,
speed of particle in positive x-direction, u = 97 m/s
speed of particle in opposite direction, v = -42 m/s
time = 2.9 s
average acceleration, a = ?
now,
![a = \dfrac{-42-97}{2.9}](https://tex.z-dn.net/?f=a%20%3D%20%5Cdfrac%7B-42-97%7D%7B2.9%7D)
![a = \dfrac{-139}{2.9}](https://tex.z-dn.net/?f=a%20%3D%20%5Cdfrac%7B-139%7D%7B2.9%7D)
a = -47.93 m/s²
Hence, the average acceleration is equal to -47.93 m/s²