Study questions platform-wide or filter by specific tests with correct answers revealed.
Cutting department data — Budget: Overheads $40,000; Machine hours 2,800; Labour hours 2,600. Actual: Overheads $44,000; Machine hours 2,700; Labour hours 2,900. What OAR would have been used?
Option C ($15.38 per labour hour) is correct.
OAR uses budgeted figures: $40,000 ÷ 2,600 budgeted labour hours = $15.38 per labour hour.
Labour hours are the more appropriate basis as actual labour hours (2,900) exceed actual machine hours (2,700), indicating a labour-intensive department.
Initially, beam A's plane is parallel to the polaroid ($\theta_A = 0^\circ$) and B's is perpendicular ($\theta_B = 90^\circ$).
After $30^\circ$ rotation, the angles become $30^\circ$ and $60^\circ$ respectively.
Equal brightness means: $I_A \cos^2(30^\circ) = I_B \cos^2(60^\circ)$
$I_A \left( \frac{\sqrt{3}}{2} \right)^2 = I_B \left( \frac{1}{2} \right)^2 \implies I_A \cdot \frac{3}{4} = I_B \cdot \frac{1}{4} \implies \frac{I_A}{I_B} = \frac{1}{3}$.
A particle starts from the origin (0, 0) and moves in a straight line. At a later time its position is $(\sqrt{3},\,3)$. The angle the path makes with the positive $x$-axis is:
$\tan\alpha = \dfrac{y}{x} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$
$\alpha = \tan^{-1}(\sqrt{3}) = \mathbf{60°}$
Sign in to join the conversation and share your thoughts.
Log In to Comment