The relationship between resistance R and resistivity
![\rho](https://tex.z-dn.net/?f=%5Crho)
is
![R= \frac{\rho L}{A}](https://tex.z-dn.net/?f=R%3D%20%5Cfrac%7B%5Crho%20L%7D%7BA%7D%20)
(1)
where L is the length of the wire and A is the cross-sectional area.
In our problem, the radius of the wire is half the diameter: r=1 mm=0.001 m, so the cross-sectional area is
![A=\pi r^2 = \pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%20%3D%20%5Cpi%20%280.001%20m%29%5E2%3D3.14%20%5Ccdot%2010%5E%7B-6%7D%20m%5E2)
The length of the wire is L=20 m and the resistance is
![R=0.25 \Omega](https://tex.z-dn.net/?f=R%3D0.25%20%5COmega)
.
By re-arranging equation (1), we can find the resistivity of the wire:
1) 30 volts is correct
2) Rt = (4x6)/(4+6)=2.4 Ω
3) P = I^2 x R = 36 W
we use AC because it's easy to transmit via 3 phase on cables and easy to step up and down
The problem about describes a perfectly inelastic collision. We are tasked to find the initial velocity of an object having a mass of 6 kg moving due west. It is given in the problem that after collision the cart sticks together and it stops. Thus, the final mass is the sum of the two cart and the final velocity is zero. For a perfectly inelastic collision,
m1v1-m2v2=vf(m1+m2)
By Substitution,
3(4)-6(v2)=0
6v2=12
v2=2
Therefor, the initial velocity if a 6 kg cart is 2 m/s
Answer: This will be based on how much strength she put into the rock when throwing it
Explanation: