An aluminum wire of length 1.0 meter has a resistance of 9.0x10^-3 ohm. If the wire were cut into two equal lengths, each length would have a resistance of

Each length of wire would have a resistance of 4.5 × 10⁻³ Ω.

Since resistance R = ρl/A where

Since ρ and A are constant for the aluminum wire, R ∝ l

So. R₂/R₁ = l₂/l₁ where

Now, since the length of the wire is cut into two, l₂ = l₁/2

So, R₂/R₁ = l₁/2 ÷ l₁

R₂/R₁ = 1/2

R₂ = R₁/2

R₂ = 9.0 × 10⁻³ Ω/2

R₂ = 4.5 × 10⁻³ Ω

So, each length of wire would have a resistance of 4.5 × 10⁻³ Ω.

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since resistance depends on length, then the resistance will be shared equally since it was cut equally therefore the resistance is

4.5*10^-3 ohm

Explanation:

continuous conducting path for current, power supply

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