See the processed text for the questions.

Steps to Solve

1. Question 14: Convert velocities to m/s Convert the initial velocity from 36 km/h to m/s: $v_i = 36 \frac{km}{h} \times \frac{1000 m}{1 km} \times \frac{1 h}{3600 s} = 10 m/s$ Convert the final velocity from 54 km/h to m/s: $v_f = 54 \frac{km}{h} \times \frac{1000 m}{1 km} \times \frac{1 h}{3600 s} = 15 m/s$

2. Question 14: Use the kinematic equation to find the time Use the equation $s = \frac{(v_i + v_f)}{2} t$, where $s$ is the distance, $v_i$ is the initial velocity, $v_f$ is the final velocity, and $t$ is the time. Plug in the values: $125 = \frac{(10 + 15)}{2} t$ Simplify: $125 = \frac{25}{2} t$

3. Question 14: Solve for time Solve for $t$: $t = \frac{125 \times 2}{25} = \frac{250}{25} = 10 s$

4. Question 15: Convert velocity to m/s Convert the speed from 720 km/h to m/s: $v = 720 \frac{km}{h} \times \frac{1000 m}{1 km} \times \frac{1 h}{3600 s} = 200 m/s$

5. Question 15: Use the formula for the radius of a banked turn The formula for the radius of a banked turn is $r = \frac{v^2}{g \tan{\theta}}$, where $v$ is the speed, $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$), and $\theta$ is the banking angle. Plug in the values: $r = \frac{(200)^2}{9.8 \tan{15^\circ}}$

6. Question 15: Calculate the radius $r = \frac{40000}{9.8 \times 0.2679} \approx \frac{40000}{2.625} \approx 15238 m = 15.238 km \approx 15 km$

7. Question 16: Recall Kepler's Third Law Kepler's Third Law states that the square of the orbital period ($T^2$) is proportional to the cube of the semi-major axis (which is approximately the radius $r$ for a circular orbit). Therefore, $T^2 \propto r^3$

8. Question 17: Recall Escape Velocity The escape velocity is independent of the mass of the object being projected. It only depends on the mass and radius of the Earth.

9. Question 18: Define Orbital Period The time taken to complete a revolution around a planet is called the orbital period.

10. Question 19: Use the kinetic energy formula to find velocity The kinetic energy formula is $KE = \frac{1}{2}mv^2$, where $KE$ is kinetic energy, $m$ is mass, and $v$ is velocity. Convert mass from grams to kilograms: $m = 200 g = 0.2 kg$ Plug in the values: $10 = \frac{1}{2}(0.2)v^2$ Solve for $v^2$: $v^2 = \frac{10 \times 2}{0.2} = \frac{20}{0.2} = 100$ Solve for $v$: $v = \sqrt{100} = 10 m/s$

11. Question 19: Calculate momentum Momentum $p = mv$, where $m$ is mass and $v$ is velocity. $p = (0.2 kg)(10 m/s) = 2 kg \cdot m/s$

12. Question 20: Apply Impulse-Momentum Theorem The impulse-momentum theorem states that the impulse (force x time) is equal to the change in momentum: $F \Delta t = \Delta p = p_f - p_i$, where $F$ is the force, $\Delta t$ is the time interval, $p_f$ is the final momentum, and $p_i$ is the initial momentum. Plug in the values: $(30 N)(3 s) = p_f - 10 kg \cdot m/s$ $90 kg \cdot m/s = p_f - 10 kg \cdot m/s$

13. Question 20: Solve for the final momentum $p_f = 90 kg \cdot m/s + 10 kg \cdot m/s = 100 kg \cdot m/s$

14. Question 21: Recall Kepler's First Law Kepler's First Law states that the orbit of a planet around the Sun is an ellipse with the Sun at one of the two foci.