Discussion
Ram rise at Mach 0.5 in air at 250 K, using \( \Delta T = 0.2 M^2 T_s \), is about
Answer: A
Correct answer: 12.5 K
\[ \Delta T = 0.2M^2T_s = 0.2 \times 0.5^2 \times 250 = 0.2 \times 0.25 \times 250 = 12.5\ \text{K.} \] Note the square: doubling the Mach number quadruples the rise, so at Mach 1 in the same air the rise would be 50 K. This is why the ram correction is negligible at low speed and dominant at high speed.
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