the given equation $r = \\frac{0.43l}{a}$ relates the resistance $r$, in ohms, of a wire to its length $l$…

the given equation $r = \\frac{0.43l}{a}$ relates the resistance $r$, in ohms, of a wire to its length $l$, in meters, and its cross - sectional area $a$, in square meters. which equation correctly expresses the length, in meters, of the wire in terms of its resistance, in ohms, and its cross - sectional area, in square meters?\na. $l = \\frac{r - a}{0.43}$\nb. $l = \\frac{r+a}{0.43}$\nc. $l = \\frac{0.43r}{a}$\nd. $l = \\frac{ar}{0.43}$

the given equation $r = \\frac{0.43l}{a}$ relates the resistance $r$, in ohms, of a wire to its length $l$, in meters, and its cross - sectional area $a$, in square meters. which equation correctly expresses the length, in meters, of the wire in terms of its resistance, in ohms, and its cross - sectional area, in square meters?\na. $l = \\frac{r - a}{0.43}$\nb. $l = \\frac{r+a}{0.43}$\nc. $l = \\frac{0.43r}{a}$\nd. $l = \\frac{ar}{0.43}$

Answer

Explanation:

Step1: Isolate (l) in the equation (R = \frac{0.43l}{A}).

Multiply both sides by (A): (RA=0.43l).

Step2: Solve for (l).

Divide both sides by (0.43): (l=\frac{RA}{0.43}).

Answer:

C. (l = \frac{RA}{0.43})