Physics Study Guide: Work, Energy, Power, Thermal Physics, Electrostatics, and Circuits

Study Guide - Smart Notes

Tailored notes based on your materials, expanded with key definitions, examples, and context.

Q1. Work by a Constant Force

Background

Topic: Work and Energy

This question tests your understanding of how to calculate the work done by a constant force and how work relates to changes in kinetic energy and speed.

Key Terms and Formulas

Work ($W$): $W = F d \cos\theta$
Kinetic Energy ($KE$): $KE = \frac{1}{2} m v^2$
Work-Energy Theorem: $W_{\text{net}} = \Delta KE$

Step-by-Step Guidance

For part (a), identify the force ($F = 12\,\text{N}$), displacement ($d = 3.5\,\text{m}$), and angle ($\theta = 0^\circ$ since the force is horizontal and so is the displacement).
Plug these values into the work formula: $W = F d \cos\theta$.
For part (b), use the work-energy theorem: the work done by the force equals the change in kinetic energy, since the box starts from rest and there is no friction.
Set $W = \Delta KE = \frac{1}{2} m v_f^2 - 0$ and solve for the final speed $v_f$.

Try solving on your own before revealing the answer!

Final Answer:

(a) $W = 42\,\text{J}$

(b) $v_f = 4.6\,\text{m/s}$

We calculated work using $W = Fd$ and then used the work-energy theorem to find the final speed.

Q2. Work Against Friction

Background

Topic: Work and Friction

This question tests your ability to calculate the work done by friction and by an applied force when moving an object at constant speed.

Key Terms and Formulas

Kinetic friction force: $f_k = \mu_k N$
Work: $W = F d \cos\theta$
At constant speed, applied force equals friction force.

Step-by-Step Guidance

Calculate the normal force: $N = mg$ (since the surface is horizontal).
Find the friction force: $f_k = \mu_k N$.
For part (a), calculate the work done by friction: $W_{\text{friction}} = f_k \times d \times \cos 180^\circ$ (since friction acts opposite to displacement).
For part (b), the student must apply a force equal to $f_k$ over the distance $d$ in the direction of motion. Calculate the work done by the student: $W_{\text{student}} = f_k \times d$.

Try solving on your own before revealing the answer!

Final Answer:

(a) $W_{\text{friction}} = -235\,\text{J}$

(b) $W_{\text{student}} = 235\,\text{J}$

The negative sign for friction indicates work is done against the direction of motion.

Q3. Gravitational Potential Energy

Background

Topic: Work and Gravitational Potential Energy

This question tests your understanding of the work done against gravity and the concept of gravitational potential energy.

Key Terms and Formulas

Work done lifting: $W = mgh$
Gravitational potential energy: $U_g = mgh$

Step-by-Step Guidance

Identify the mass ($m = 2.0\,\text{kg}$), height ($h = 1.8\,\text{m}$), and acceleration due to gravity ($g = 9.8\,\text{m/s}^2$).
For part (a), calculate the work done by the person: $W = mgh$.
For part (b), the gravitational potential energy gained is also $U_g = mgh$.

Try solving on your own before revealing the answer!

Final Answer:

(a) $W = 35.3\,\text{J}$

(b) $U_g = 35.3\,\text{J}$

The work done lifting equals the increase in gravitational potential energy.

Q4. Kinetic Energy and Speed

Background

Topic: Kinetic Energy

This question tests your ability to relate kinetic energy to the speed of an object and understand how kinetic energy changes with speed.

Key Terms and Formulas

Kinetic energy: $KE = \frac{1}{2} m v^2$

Step-by-Step Guidance

For part (a), set $KE = \frac{1}{2} m v^2$ and solve for $v$ using the given values ($KE = 3.0 \times 10^5\,\text{J}$, $m = 1500\,\text{kg}$).
For part (b), recall that kinetic energy depends on the square of the speed. If speed doubles, $KE$ changes by a factor of $(2)^2$.

Try solving on your own before revealing the answer!

Final Answer:

(a) $v = 20\,\text{m/s}$

(b) The kinetic energy increases by a factor of 4.

Doubling the speed quadruples the kinetic energy because $KE \propto v^2$.

Q5. Power Output

Background

Topic: Power and Work

This question tests your understanding of the relationship between power, force, velocity, and work done over time.

Key Terms and Formulas

Power: $P = Fv$ (for constant speed)
Work: $W = Fd$
Weight: $F = mg$ (force needed to lift at constant speed)

Step-by-Step Guidance

Calculate the force required to lift the mass: $F = mg$.
For part (a), use $P = Fv$ with the calculated force and given speed.
For part (b), calculate the work done in 12 s: $W = P \times t$ or $W = F \times d$ (where $d = v \times t$).

Try solving on your own before revealing the answer!

Final Answer:

(a) $P = 184\,\text{W}$

(b) $W = 2200\,\text{J}$

We used the power formula for constant speed and calculated work as force times distance.

Q6. Conservation of Mechanical Energy — Falling Object

Background

Topic: Conservation of Energy

This question tests your understanding of how gravitational potential energy converts to kinetic energy as an object falls.

Key Terms and Formulas

Potential energy: $U_g = mgh$
Kinetic energy: $KE = \frac{1}{2} m v^2$
Conservation of energy: $U_{g,\text{initial}} = KE_{\text{final}}$ (if no energy is lost)

Step-by-Step Guidance

Calculate the initial gravitational potential energy: $U_g = mgh$.
Set $U_g = KE$ just before impact, since the ball starts from rest and all potential energy converts to kinetic energy.
Solve for the final speed $v$ using $KE = \frac{1}{2} m v^2$.
For part (b), use the value of $v$ to find the kinetic energy at impact.

Try solving on your own before revealing the answer!

Final Answer:

(a) $v = 7.0\,\text{m/s}$

(b) $KE = 3.7\,\text{J}$

All potential energy is converted to kinetic energy just before impact.

Q7. Work–Energy Principle

Background

Topic: Work-Energy Theorem

This question tests your ability to apply the work-energy principle to relate work done to changes in kinetic energy and speed.

Key Terms and Formulas

Work-Energy Theorem: $W = \Delta KE = KE_f - KE_i$
Kinetic energy: $KE = \frac{1}{2} m v^2$

Step-by-Step Guidance

Calculate the initial kinetic energy: $KE_i = \frac{1}{2} m v_i^2$.
Add the work done to find the final kinetic energy: $KE_f = KE_i + W$.
Solve for the final speed $v_f$ using $KE_f = \frac{1}{2} m v_f^2$.
For part (b), the change in kinetic energy is simply $\Delta KE = W$.

Try solving on your own before revealing the answer!

Final Answer:

(a) $v_f = 4.2\,\text{m/s}$

(b) $\Delta KE = 600\,\text{J}$

The work done equals the change in kinetic energy.

Q8. Power and Energy Consumption

Background

Topic: Power and Energy

This question tests your understanding of the relationship between power, energy, time, and cost of electricity.

Key Terms and Formulas

Power: $P = \frac{E}{t}$
Energy: $E = P \times t$
1 kWh = $3.6 \times 10^6$ J

Step-by-Step Guidance

Convert 10 minutes to seconds: $t = 10 \times 60$ s.
Calculate energy used: $E = P \times t$.
Convert energy from joules to kilowatt-hours for part (b).
Multiply the energy in kWh by the cost per kWh to find the total cost.

Try solving on your own before revealing the answer!

Final Answer:

(a) $E = 51,000\,\text{J}$

(b) Cost $= \$0.00017$

We converted energy to kWh and multiplied by the cost per kWh.

Q9. Temperature Scales

Background

Topic: Temperature Conversion

This question tests your ability to convert between Fahrenheit, Celsius, and Kelvin temperature scales.

Key Terms and Formulas

$T_{\text{C}} = \frac{5}{9}(T_{\text{F}} - 32)$
$T_{\text{K}} = T_{\text{C}} + 273.15$

Step-by-Step Guidance

For part (a), plug $T_{\text{F}} = 86^\circ\text{F}$ into the Celsius conversion formula.
For part (b), add 273.15 to the Celsius temperature to get Kelvin.

Try solving on your own before revealing the answer!

Final Answer:

(a) $T_{\text{C}} = 30^\circ\text{C}$

(b) $T_{\text{K}} = 303.15\,\text{K}$

We used the standard temperature conversion formulas.

Q10. Heat and Temperature Change

Background

Topic: Specific Heat

This question tests your ability to calculate the heat absorbed or released when a substance changes temperature.

Key Terms and Formulas

Heat: $Q = mc\Delta T$

Step-by-Step Guidance

Identify the mass ($m = 0.50\,\text{kg}$), specific heat ($c = 900\,\text{J/(kg}^\circ\text{C})$), and temperature change ($\Delta T = 75 - 20$).
Plug these values into $Q = mc\Delta T$ to find the heat absorbed.

Try solving on your own before revealing the answer!

Final Answer:

$Q = 24,750\,\text{J}$

We calculated the heat using the specific heat formula.

Q11. Mixing Water at Different Temperatures

Background

Topic: Calorimetry

This question tests your ability to apply the principle of conservation of energy to find the final equilibrium temperature when mixing water samples at different temperatures.

Key Terms and Formulas

Heat lost = Heat gained
$m_1 c (T_f - T_1) + m_2 c (T_f - T_2) = 0$

Step-by-Step Guidance

Let $T_f$ be the final temperature. Set up the equation: $m_1 (T_f - T_1) + m_2 (T_f - T_2) = 0$ (since $c$ is the same for both).
Plug in $m_1 = 200\,\text{g}$, $T_1 = 80^\circ\text{C}$, $m_2 = 300\,\text{g}$, $T_2 = 20^\circ\text{C}$.
Solve for $T_f$ algebraically before plugging in numbers.

Try solving on your own before revealing the answer!

Final Answer:

$T_f = 40^\circ\text{C}$

The final temperature is a weighted average based on the masses and initial temperatures.

Q12. Phase Change — Melting Ice

Background

Topic: Latent Heat

This question tests your ability to calculate the heat required for a phase change using the latent heat of fusion.

Key Terms and Formulas

Heat for melting: $Q = mL_f$

Step-by-Step Guidance

Identify the mass ($m = 0.25\,\text{kg}$) and latent heat of fusion ($L_f = 3.34 \times 10^5\,\text{J/kg}$).
Plug these values into $Q = mL_f$ to find the heat required.

Try solving on your own before revealing the answer!

Final Answer:

$Q = 83,500\,\text{J}$

We multiplied the mass by the latent heat of fusion.

Q13. Thermal Conduction

Background

Topic: Heat Transfer by Conduction

This question tests your ability to calculate the rate of heat transfer through a material using its thermal conductivity.

Key Terms and Formulas

Rate of heat conduction: $P = \frac{kA\Delta T}{d}$

Step-by-Step Guidance

Identify the thermal conductivity ($k = 1.0\,\text{W/(m}\cdot\text{K)}$), area ($A = 1\,\text{m}^2$), thickness ($d = 0.02\,\text{m}$), and temperature difference ($\Delta T = 20 - (-10) = 30^\circ\text{C}$).
Plug these values into $P = \frac{kA\Delta T}{d}$ to find the rate of heat conduction.

Try solving on your own before revealing the answer!

Final Answer:

$P = 1,500\,\text{W}$

We used the conduction formula with the given values.

Q14. Specific Heat — Cooling

Background

Topic: Specific Heat and Cooling

This question tests your ability to calculate the heat lost by a substance as it cools.

Key Terms and Formulas

Heat lost: $Q = mc\Delta T$ (where $\Delta T$ is negative for cooling)

Step-by-Step Guidance

Identify the mass ($m = 1.2\,\text{kg}$), specific heat ($c = 4186\,\text{J/(kg}^\circ\text{C})$), and temperature change ($\Delta T = 65 - 95$).
Plug these values into $Q = mc\Delta T$ to find the heat lost.

Try solving on your own before revealing the answer!

Final Answer:

$Q = -150,700\,\text{J}$

The negative sign indicates heat is lost by the soup.

Q15. Linear Expansion of a Metal Rod

Background

Topic: Thermal Expansion

This question tests your ability to calculate the change in length of a rod due to temperature change using the coefficient of linear expansion.

Key Terms and Formulas

Change in length: $\Delta L = \alpha L_0 \Delta T$

Step-by-Step Guidance

Identify the original length ($L_0 = 2.00\,\text{m}$), coefficient of linear expansion ($\alpha = 1.2 \times 10^{-5}\,\text{C}^{-1}$), and temperature change ($\Delta T = 80 - 20$).
Plug these values into $\Delta L = \alpha L_0 \Delta T$ to find the change in length.

Try solving on your own before revealing the answer!

Final Answer:

$\Delta L = 0.00144\,\text{m}$

The rod expands by this amount when heated from 20°C to 80°C.

Q16. Expansion Fit Problem

Background

Topic: Thermal Expansion (Engineering Application)

This question tests your ability to use the linear expansion formula to determine the temperature needed for a metal ring to expand enough to fit over a shaft.

Key Terms and Formulas

Change in diameter: $\Delta D = \alpha D_0 \Delta T$

Step-by-Step Guidance

Set $\Delta D = D_{\text{shaft}} - D_{\text{ring, initial}}$.
Use $\Delta D = \alpha D_0 \Delta T$ to solve for $\Delta T$.
Add $\Delta T$ to the initial temperature to find the required temperature.

Try solving on your own before revealing the answer!

Final Answer:

Required temperature $= 59^\circ\text{C}$

The ring must be heated to this temperature for the shaft to fit.

Q17. Volume Expansion of a Liquid

Background

Topic: Volume Expansion

This question tests your ability to calculate the change in volume of a liquid due to temperature change using the coefficient of volume expansion.

Key Terms and Formulas

Change in volume: $\Delta V = \beta V_0 \Delta T$
Final volume: $V = V_0 + \Delta V$

Step-by-Step Guidance

Identify the initial volume ($V_0 = 0.500\,\text{L}$), coefficient of volume expansion ($\beta = 9.5 \times 10^{-4}\,\text{C}^{-1}$), and temperature change ($\Delta T = 35 - 10$).
Calculate $\Delta V = \beta V_0 \Delta T$.
Add $\Delta V$ to $V_0$ to find the final volume.

Try solving on your own before revealing the answer!

Final Answer:

$V = 0.506\,\text{L}$

The gasoline expands as temperature increases.

Q18. Charge and Number of Electrons

Background

Topic: Quantization of Charge

This question tests your ability to relate the total charge on an object to the number of excess electrons.

Key Terms and Formulas

Charge of one electron: $e = 1.6 \times 10^{-19}\,\text{C}$
Number of electrons: $n = \frac{|q|}{e}$

Step-by-Step Guidance

Identify the total charge ($q = -4.8 \times 10^{-19}\,\text{C}$).
Calculate the number of excess electrons: $n = \frac{|q|}{e}$.

Try solving on your own before revealing the answer!

Final Answer:

$n = 3.0 \times 10^0$

We divided the total charge by the elementary charge.

Q19. Coulomb’s Law — Two Point Charges

Background

Topic: Electrostatics

This question tests your ability to calculate the electrostatic force between two point charges using Coulomb’s law.

Key Terms and Formulas

Coulomb’s law: $F = k_e \frac{|q_1 q_2|}{r^2}$
$k_e = 8.99 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$

Step-by-Step Guidance

Identify the charges ($q_1 = 2.0\,\mu\text{C}$, $q_2 = 3.0\,\mu\text{C}$) and separation ($r = 0.15\,\text{m}$).
Convert microcoulombs to coulombs: $1\,\mu\text{C} = 1 \times 10^{-6}\,\text{C}$.
Plug values into Coulomb’s law to find the force magnitude.

Try solving on your own before revealing the answer!

Final Answer:

$F = 2.4\,\text{N}$

We used Coulomb’s law with the correct units.

Q20. Attractive vs. Repulsive Force

Background

Topic: Electrostatics — Nature of Forces

This question tests your understanding of the direction (attractive or repulsive) and magnitude of the force between charges of opposite sign.

Key Terms and Formulas

Coulomb’s law: $F = k_e \frac{|q_1 q_2|}{r^2}$
Like charges repel, unlike charges attract.

Step-by-Step Guidance

Identify the charges ($+5.0\,\mu\text{C}$ and $-1.0\,\mu\text{C}$) and separation ($r = 0.25\,\text{m}$).
Determine the nature of the force (attractive or repulsive) based on the signs of the charges.
Calculate the magnitude using Coulomb’s law.

Try solving on your own before revealing the answer!

Final Answer:

The force is attractive; $F = 0.72\,\text{N}$

Opposite charges attract, and we used Coulomb’s law for the magnitude.

Q21. Coulomb’s Law — Force Between Two Charges

Background

Topic: Electrostatics

This question tests your ability to calculate both the magnitude and direction of the force between two point charges.

Key Terms and Formulas

Coulomb’s law: $F = k_e \frac{|q_1 q_2|}{r^2}$
Direction: Attractive if charges are opposite, repulsive if same sign.

Step-by-Step Guidance

Identify the charges ($q_1 = +3.0\,\mu\text{C}$, $q_2 = -5.0\,\mu\text{C}$) and separation ($r = 0.25\,\text{m}$).
Calculate the magnitude using Coulomb’s law.
Determine the direction: the force is attractive, so each charge pulls toward the other.

Try solving on your own before revealing the answer!

Final Answer:

$F = 2.2\,\text{N}$, attractive

The force acts along the line joining the charges, pulling them together.

Q22. Electric Field of a Point Charge

Background

Topic: Electric Field

This question tests your ability to calculate the electric field produced by a point charge at a given distance.

Key Terms and Formulas

Electric field: $E = k_e \frac{|q|}{r^2}$

Step-by-Step Guidance

Identify the charge ($q = 8.0\,\text{nC} = 8.0 \times 10^{-9}\,\text{C}$) and distance ($r = 0.40\,\text{m}$).
Plug values into $E = k_e \frac{|q|}{r^2}$ to find the electric field magnitude.

Try solving on your own before revealing the answer!

Final Answer:

$E = 450\,\text{N/C}$

We used the electric field formula for a point charge.

Q23. Force on a Charge in an Electric Field

Background

Topic: Electric Force

This question tests your ability to calculate the force on a charge placed in a uniform electric field and determine its direction.

Key Terms and Formulas

Force: $F = qE$
Direction: For a negative charge, force is opposite to the field direction.

Step-by-Step Guidance

Identify the charge ($q = -3.0\,\text{nC} = -3.0 \times 10^{-9}\,\text{C}$) and electric field ($E = 2.5 \times 10^5\,\text{N/C}$).
Calculate the magnitude: $F = |q|E$.
Determine the direction: since the charge is negative, the force is opposite to the field direction.

Try solving on your own before revealing the answer!

Final Answer:

$F = 0.00075\,\text{N}$, to the left

The force is opposite the field direction for a negative charge.

Q24. Electric Field of a Point Charge

Background

Topic: Electric Field

This question tests your ability to calculate the electric field at a distance from a point charge.

Key Terms and Formulas

Electric field: $E = k_e \frac{|q|}{r^2}$

Step-by-Step Guidance

Identify the charge ($q = 6.0\,\text{nC} = 6.0 \times 10^{-9}\,\text{C}$) and distance ($r = 0.20\,\text{m}$).
Plug values into $E = k_e \frac{|q|}{r^2}$ to find the electric field magnitude.

Try solving on your own before revealing the answer!

Final Answer:

$E = 1,350\,\text{N/C}$

We used the electric field formula for a point charge.

Q25. Direction of the Electric Field

Background

Topic: Electric Field Direction

This question tests your understanding of the direction of the electric field produced by a negative charge.

Key Terms and Formulas

The electric field points toward negative charges and away from positive charges.

Step-by-Step Guidance

At a point to the right of a negative charge, the electric field points toward the charge (to the left).

Try solving on your own before revealing the answer!

Final Answer:

The electric field points to the left (toward the negative charge).

Electric field lines point toward negative charges.

Q26. Net Electric Field From Two Charges (Simple Geometry)

Background

Topic: Superposition Principle

This question tests your ability to calculate the net electric field at a point due to two charges using the principle of superposition.

Key Terms and Formulas

Electric field from a point charge: $E = k_e \frac{q}{r^2}$
Net field: $E_{\text{net}} = E_1 + E_2$ (vector sum)

Step-by-Step Guidance

Calculate the electric field at the midpoint due to each charge separately (both are $0.5\,\text{m}$ away).
Add the fields, considering their directions (both charges are positive, so fields at the midpoint point away from each charge).

Try solving on your own before revealing the answer!

Final Answer:

$E_{\text{net}} = 1.8 \times 10^5\,\text{N/C}$, to the right

We summed the fields from both charges at the midpoint.

Q27. Equal and Opposite Charges

Background

Topic: Electrostatics

This question tests your ability to calculate the force between two equal and opposite charges.

Key Terms and Formulas

Coulomb’s law: $F = k_e \frac{|q_1 q_2|}{r^2}$

Step-by-Step Guidance

Identify the charges ($+3.0\,\mu\text{C}$ and $-3.0\,\mu\text{C}$) and separation ($r = 0.60\,\text{m}$).
Calculate the magnitude using Coulomb’s law.

Try solving on your own before revealing the answer!

Final Answer:

$F = 0.225\,\text{N}$

The force is attractive and equal in magnitude for both charges.

Q28. Charge of a Sphere

Background

Topic: Quantization of Charge

This question tests your ability to determine the number of electrons removed from a sphere given its net positive charge.

Key Terms and Formulas

Charge of one electron: $e = 1.6 \times 10^{-19}\,\text{C}$
Number of electrons: $n = \frac{q}{e}$

Step-by-Step Guidance

Identify the net charge ($q = 1.6 \times 10^{-19}\,\text{C}$).
Calculate the number of electrons removed: $n = \frac{q}{e}$.

Try solving on your own before revealing the answer!

Final Answer:

$n = 1$

One electron was removed to give the sphere this net charge.

Q29. Electric Field From Two Symmetric Charges

Background

Topic: Superposition Principle

This question tests your ability to calculate the net electric field at the midpoint between two identical charges.

Key Terms and Formulas

Electric field from a point charge: $E = k_e \frac{q}{r^2}$
Net field: $E_{\text{net}} = 2E$ (since both fields point in the same direction at the midpoint)

Step-by-Step Guidance

Calculate the field at the midpoint due to one charge ($r = 0.20\,\text{m}$ from each charge).
Double the field to get the net field, since both charges are identical and fields add.

Try solving on your own before revealing the answer!

Final Answer:

$E_{\text{net}} = 2.25 \times 10^3\,\text{N/C}$

The fields from both charges add at the midpoint.

Q30. Ohm’s Law — Solving for Current

Background

Topic: Ohm’s Law

This question tests your ability to use Ohm’s law to find the current through a resistor given voltage and resistance.

Key Terms and Formulas

Ohm’s law: $I = \frac{V}{R}$

Step-by-Step Guidance

Identify the voltage ($V = 12\,\text{V}$) and resistance ($R = 4.0\,\Omega$).
Plug values into $I = \frac{V}{R}$ to find the current.

Try solving on your own before revealing the answer!

Final Answer:

$I = 3.0\,\text{A}$

We divided voltage by resistance to find the current.

Q31. Ohm’s Law — Solving for Resistance

Background

Topic: Ohm’s Law

This question tests your ability to use Ohm’s law to find resistance given current and voltage.

Key Terms and Formulas

Ohm’s law: $R = \frac{V}{I}$

Step-by-Step Guidance

Identify the voltage ($V = 9.0\,\text{V}$) and current ($I = 0.50\,\text{A}$).
Plug values into $R = \frac{V}{I}$ to find the resistance.

Try solving on your own before revealing the answer!

Final Answer:

$R = 18\,\Omega$

We divided voltage by current to find resistance.

Q32. Voltage Drop Across a Resistor

Background

Topic: Ohm’s Law

This question tests your ability to calculate the voltage drop across a resistor given current and resistance.

Key Terms and Formulas

Ohm’s law: $V = IR$

Step-by-Step Guidance

Identify the current ($I = 2.0\,\text{A}$) and resistance ($R = 3.0\,\Omega$).
Plug values into $V = IR$ to find the voltage drop.

Try solving on your own before revealing the answer!

Final Answer:

$V = 6.0\,\text{V}$

We multiplied current by resistance to find the voltage drop.

Q33. Power in a Resistor

Background

Topic: Electric Power

This question tests your ability to calculate the power dissipated in a resistor given current and resistance.

Key Terms and Formulas

Power: $P = I^2 R$
Alternatively, $P = VI$

Step-by-Step Guidance

Identify the current ($I = 0.75\,\text{A}$) and resistance ($R = 20\,\Omega$).
Plug values into $P = I^2 R$ to find the power dissipated.

Try solving on your own before revealing the answer!

Final Answer:

$P = 11.25\,\text{W}$

We used the power formula for resistors.

Q34. Resistivity and Wire Resistance

Background

Topic: Resistivity

This question tests your ability to calculate the resistance of a wire using its resistivity, length, and cross-sectional area.

Key Terms and Formulas

Resistance: $R = \rho \frac{L}{A}$
$\rho$ = resistivity, $L$ = length, $A$ = area

Step-by-Step Guidance

Identify the resistivity ($\rho = 1.68 \times 10^{-8}\,\Omega\cdot\text{m}$), length ($L = 5.0\,\text{m}$), and area ($A = 2.0 \times 10^{-6}\,\text{m}^2$).
Plug values into $R = \rho \frac{L}{A}$ to find the resistance.

Try solving on your own before revealing the answer!

Final Answer:

$R = 0.042\,\Omega$

We used the resistivity formula for a wire.