I don't know I guess its the plate tectonics
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.
The relationship between the resistance and the resistivity of a wire is
![\rho = \frac{AR}{L}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%20%5Cfrac%7BAR%7D%7BL%7D%20)
where
![\rho](https://tex.z-dn.net/?f=%5Crho)
is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by
![A=\pi r^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2)
where r is the radius of the wire. Substituting in the previous equation ,we find
![\rho = \frac{\pi r^2 R}{L}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%20%5Cfrac%7B%5Cpi%20r%5E2%20R%7D%7BL%7D%20)
For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
![\rho' = \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4 \frac{\pi r^2 R}{L} = 4 \rho](https://tex.z-dn.net/?f=%5Crho%27%20%3D%20%20%5Cfrac%7B%5Cpi%20r%27%5E2%20R%7D%7BL%27%7D%3D%20%5Cfrac%7B%5Cpi%20%282r%29%5E2%20R%7D%7BL%7D%3D4%20%20%5Cfrac%7B%5Cpi%20r%5E2%20R%7D%7BL%7D%20%20%20%3D%204%20%5Crho)
Therefore, the new resistivity must be 4 times the original one.
Answer:
A
Explanation:
Hooke's law! F(spring)=-kx
There's no tricky square law here. The spring constant doesn't change, only x (distance stretched) changes. Therefore, if distance is halved, Force will be halved.
The pet store would be the reference point because it is where he started and it will not move. Hope this helped.