VCAA Physics How do physicists explain motion in two dimensions?

This option is incorrect. It is close to the value of the circumference ($C = vT = 2.4 \text{ m}$) or the result of dividing speed by $\pi$, rather than solving for the radius using $2\pi$.

This option is incorrect and results from a calculation error or misapplication of the circular motion variables.

This option is incorrect. It results from incorrectly rearranging the formula as $r = \frac{2\pi v}{T}$ instead of dividing by $2\pi$.

Incorrect. This value is obtained by multiplying Earth's escape velocity by $3/4$, incorrectly assuming escape velocity is proportional to $R/M$.

Incorrect. This value is obtained by multiplying Earth's escape velocity by $\sqrt{3/4}$, incorrectly assuming escape velocity is proportional to $\sqrt{R/M}$.

Incorrect. This value is obtained by multiplying Earth's escape velocity by $4/3$, incorrectly assuming escape velocity is proportional to $M/R$ instead of $\sqrt{M/R}$.

This incorrect value results from calculating the sine of the angle in radians ($50.0 \sin(40 \text{ rad}) \approx 37.3 \, \text{N}$) rather than degrees.

This represents the perpendicular component of gravity ($F_g \cos(40.0^\circ) \approx 38.3 \, \text{N}$), which is balanced by the normal force and does not contribute to the net force down the incline.

This value corresponds to $F_g \tan(40.0^\circ) \approx 42.0 \, \text{N}$, which is an incorrect trigonometric relationship for finding the parallel force component.

The kinematic equation for an object falling from rest is $v^2 = 2gs$, meaning $v$ is proportional to $\sqrt{s}$. A graph of $v$ versus $s$ would result in a curve, not a straight line.

Based on the equation $v^2 = 2gs$, $v^2$ is directly proportional to $s$, not $\sqrt{s}$. Plotting $v^2$ versus $\sqrt{s}$ would result in a quadratic curve.

Since $v$ is proportional to $s^{1/2}$, $\sqrt{v}$ would be proportional to $s^{1/4}$. Plotting $\sqrt{v}$ versus $s$ would not produce a straight line.

This assumes the orbital radius is inversely proportional to the square of the satellite's mass. However, orbital speed and radius are completely independent of the satellite's mass.

This incorrectly assumes the orbital radius is directly proportional to the satellite's mass. The mass of the orbiting object cancels out when equating gravitational and centripetal forces.

This incorrectly assumes the orbital radius is proportional to the square of the satellite's mass. A satellite's mass has no effect on its orbital radius for a given constant speed.

This describes the direction of the apparent "centrifugal" force (a pseudo-force); the actual centripetal force must pull inward to keep the object on a curved path.

A force acting opposite to the direction of motion acts as a braking force that slows the object down, rather than causing the perpendicular acceleration required for circular motion.

A tangential force acts parallel to the velocity and changes the object's speed, whereas centripetal force acts perpendicular to the velocity to change the direction.

This value is extremely small (less than a minute) and is physically impossible for a satellite in such a large orbit.

This value corresponds to the orbital period in seconds ($1.64 \times 10^6$ s), but the question asks for the answer in hours.

This value is significantly too large and likely results from a calculation error involving the powers of ten or failing to take the square root.

This option is incorrect; it may result from a calculation error or using an incorrect fraction for the vertical component.

This answer incorrectly uses the horizontal component (cosine) instead of the vertical component (sine), calculating $\frac{v^2 \cos^2(\theta)}{2g}$.

This option ignores the launch angle and calculates the height as if the projectile were fired straight up ($90^\circ$) using $\frac{v^2}{2g}$.

This answer ignores the frictional force. It calculates the acceleration based solely on the component of gravity acting down the slope ($a = g \sin 35^\circ \approx 5.6\text{ m s}^{-2}$).

This calculation incorrectly uses the cosine function for the parallel component of gravity ($mg \cos 35^\circ$) instead of the sine function. The cosine component represents the normal force, not the force pulling the object down the slope.

This result incorrectly adds the frictional force to the gravitational component instead of subtracting it. Friction always opposes the direction of motion, so it must be subtracted from the driving force.