Here are the solutions to the questions: Question 17: Step 1: State the formula for the power of a lens. The power of a lens $P$ is the reciprocal of its focal length $f$ in meters. For a converging lens, the focal length is positive. $$ P = \frac{1}{f} $$ Step 2: Convert the focal length to meters. Given $f = 20 \, \text{cm}$. $$ f = 20 \, \text{cm} = 0.20 \, \text{m} $$ Step 3: Calculate the power of the lens. $$ P = \frac{1}{0.20 \, \text{m}} = 5 \, \text{D} $$ Since it is a converging lens, the power is positive. $$ P = +5.0 \, \text{D} $$ The final answer is $\boxed{\text{B. } +5.0 \, \text{D}}$. Question 18: Step 1: State the relationship between the focal length and radius of curvature for a spherical mirror. The focal length $f$ of a spherical mirror is half its radius of curvature $R$. For a convex mirror, the focal length is negative. $$ f = \frac{R}{2} $$ Step 2: Substitute the given radius of curvature. Given $R = 2 \, \text{m}$. $$ f = \frac{2 \, \text{m}}{2} = 1 \, \text{m} $$ Step 3: Apply the sign convention for a convex mirror. For a convex mirror, the focal length is negative. $$ f = -1 \, \text{m} $$ The final answer is $\boxed{\text{C. } -1 \, \text{m}}$. Question 19: Ohm's law states that the voltage ($V$) across a conductor is directly proportional to the current ($I$) flowing through it, with the constant of proportionality being the resistance ($R$). $$ V = IR $$ The final answer is $\boxed{\text{A. } V = IR}$. Question 20: The resistance $R$ of a wire is given by the formula: $$ R = \rho \frac{L}{A} $$ where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area. From this formula, resistance is directly proportional to length. Therefore, an increase in length will increase the resistance of the wire. The final answer is $\boxed{\text{D. increase in length}}$. Question 21: The unit of electric current is the ampere. The final answer is $\boxed{\text{B. ampere}}$. What's next?