VCE Physics worked exam problems by Area of Study quiz

$20 \cos(30^\circ) \approx 17.3$ m/s

$20 \tan(30^\circ) \approx 11.5$ m/s

$20 \sin(30^\circ) = 10$ m/s, since the angle is measured from the vertical

$20$ m/s, because horizontal speed equals the full launch speed

$\frac{v}{r}$ directed along the tangent to the circle

$\frac{v^2}{r}$ directed toward the centre

$v^2 r$ directed away from the centre of the circle

Zero, since the speed is constant in uniform circular motion

Both masses move forward together at half the incoming speed, sharing the momentum equally

Both masses rebound backward at the original incoming speed

Incoming mass stops; struck mass moves forward at incoming speed

The incoming mass keeps its full speed and passes straight through the struck mass

$U$ increases by $10$ J

$U$ increases by $49$ J

$U$ decreases

$U$ increases by $98$ J

Roughly $90$ minutes

About $24$ hours, the same as a geostationary orbit

About $1$ year, matching Earth's orbit of the Sun

Around $10$ seconds for such a low altitude

The product of field and area, $B$ times $A$, regardless of how it changes

Negative rate of change of magnetic flux: $-\frac{d\Phi}{dt}$

The magnetic flux itself, $\Phi = BA \cos\theta$, threading the loop

Zero in all closed loops because flux cannot drive a current

$24$ V

$240$ V

$2400$ V

$24000$ V

$v = f + \lambda$

$v = \frac{f}{\lambda}$

$v = \frac{\lambda}{f}$

$v = f \lambda$

Increases fringe spacing

Decreases fringe spacing

Has no effect

Removes fringes

$W = \frac{h}{f_0}$

$W = h f_0$

$W = mc^2$

$W = KE_{\max}$