Question

question 3 of 10
a skydiver finds that she speeds up when she holds her arms close to her body. what does this do?

a. it reduces the force of gravity.
b. it reduces the force of air resistance.
c. it increases the force of air resistance.
d. it increases the force of gravity.

Explanation:

• The force of gravity \(F = mg\) (where \(m\) is mass and \(g\) is acceleration due to gravity). Mass \(m\) of the skydiver and \(g\) (approx \(9.8\ m/s^{2}\) near Earth's surface) do not change when she holds her arms close. So options A and D are incorrect.
• Air - resistance \(F_{air}\) is related to the cross - sectional area. When the skydiver holds her arms close to her body, the cross - sectional area \(A\) (the area perpendicular to the direction of motion) decreases. Using the formula for air resistance \(F_{air}=\frac{1}{2}

ho v^{2}C_{d}A\) (where \(
ho\) is air density, \(v\) is velocity, \(C_{d}\) is the drag coefficient), a decrease in \(A\) leads to a decrease in \(F_{air}\). When \(F_{air}\) (opposite to the direction of motion) decreases and the force of gravity (downward) remains the same, the net force \(F_{net}=F_{g}-F_{air}\) (downward) increases, and according to \(F = ma\), the acceleration (and thus speed over time) increases.

Answer:

B. It reduces the force of air resistance.