How long must a 0.54-mm-diameter aluminum wire be to have a 0.42 A current when connected to the terminals of a 1.5 V flashlight battery

L = 30.85 m

Explanation:

First, we calculate the resistance of the wire by using Ohm's Law:

V = IR

where,

V = Potential Difference = 1.5 V

I = Current = 0.42 A

R = Resistance of Wire = ?

Therefore,

Now, the cross-sectional area of wire will be:

Now, the resistance of the wire is given as:

where,

L = Length of Wire = ?

ρ = resistivity of aluminum = 2.65×10⁻⁸ Ohm.m

Therefore,

L = 30.85 m

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards