More mass, more inertia, less speed, more momentum because momentum is depends its mass and speed. Hope it helps
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

where

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

where r is the radius of the wire. Substituting in the previous equation ,we find

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

Therefore, the new resistivity must be 4 times the original one.
What’s the question here?
Assume the snow is uniform, and horizontal.
Given:
coefficient of kinetic friction = 0.10 = muK
weight of sled = 48 N
weight of rider = 660 N
normal force on of sled with rider = 48+660 N = 708 N = N
Force required to maintain a uniform speed
= coefficient of kinetic friction * normal force
= muK * N
= 0.10 * 708 N
=70.8 N
Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.
The first one is dependent variable
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