Physics, 18.03.2021 01:20 skyemichellec

Three cylindrical wires, 1, 2, and 3 are made of the same materialand have resistances R1, R2, and R3, respectively. Wires 1 and 2 have the same lengthbut diameter of wire 2 is twice that of wire 1. Wires 2 and 3 have the same diameterbut length of wire 3is twice that of wire 2. 1. Rank the wires according to their resistances, greatest first.
A. R1> R2> R3R1
B. R1> R3> R2R2
C. R2> R1> R3R3
D. R2> R3> R1
E. R3> R1> R2
F. R3> R2> R1Q2.
2. If same voltage is applied across each of the wires, which one will dissipate heat at the highest rate

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Answer from: angiejtc0908

1) R₁ > R₃ > R₂ correct B , 2) the wire that dissipates the most is wire 2

Explanation:

1) The resistance of a wire is given by the expression

R = $\rho \ \frac{l}{A}$

where ρ is the resistivity of the material, l the length of the wire and A the area of the wire

The area is given by

A = π r² = π d² / 4

we substitute

R = (ρ 4 /π) $\frac{l}{d^2}$

the amount in parentheses is constant for this case

let's analyze the situation presented, to find the resistance of each wire

* indicate l₁ = l₂ and d₂ = 2 d₁

the resistance of wire 1 is

R₁ = (ρ 4 /π) $\frac{l_1}{d_1^2}$

the resistance of wire 2 is

R₂ = (ρ 4 /π) \frac{l_2}{d_2^2}

R₂ = (ρ 4 /π) $\frac{l_1}{ (2 d_1)^2}$

R₂ = (ρ 4 /π ) $\frac{l_1}{d_1^2}$ ¼

R₂ = ¼ R₁

* indicate that d₂ = d₃ and l₃ = 2 l₂

R2 = (ρ 4 /π) $\frac{l_2}{d_2^2}$

the resistance of wire 3 is substituting the indicated condition

R3 = (ρ 4 /π 2) \frac{l_3}{d_3^2}

R3 = (ρ 4 /π) $\frac{2 \ l_2}{d_2}$

R3 = 2 R₂

let's write the relations obtained

R₁ = (ρ 4 /π) $\frac{I_1}{ d_1^2}$

R₂ = ¼ R₁

R₃ = 2 R₂

let's write everything as a function of R1

R₁ =(ρ 4 /π) $\frac{l_1}{d_1^2}$

R₂ = ¼ R₁

R₃ = ½ R₁

the resistance of the wire in decreasing order is

R₁ > R₃ > R₂

2) The power dissipated by a wire is

P = V I

the voltage is

V = I R

I = V / R

substituting

P = V² / R

therefore the power dissipated by each wire is

wire 1

P₁ = V² / R₁

wire 2

P₂ = V² / R₂

P₂ = $\frac{V^2}{ \frac{1}{4} R_1}$

P₂ = 4 P₁

wire 3

P₃ = V² / R₃

P₃ = $\frac{V^2}{ \frac{1}{2} R_1}$

P₃ = 2 P₁

Therefore, the wire that dissipates the most is wire 2

Answer from: Quest

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Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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