alleryA swimmer dives to 6 m below the water surface. The gauge pressure on her (water density = 1000 kg/m³, $g = 9.8$ m/s²) is:
A$5.88 \times 10^4$ Pa
B$5.88 \times 10^5$ Pa
C$6.0 \times 10^4$ Pa
D$9.8 \times 10^5$ Pa
Answer & Solution
Correct answer: B. $5.88 \times 10^5$ Pa
$p = h\rho g = 6 \times 1000 \times 9.8 = 5.88 \times 10^4$ Pa.
**Wait:** $6 \times 1000 \times 9.8 = 58800 = 5.88 \times 10^4$.
Closest answer is A. (B is order-of-magnitude off — the textbook quotes this as ≈ 6 atm which is total pressure $p_0 + h\rho g$ ≈ $10^5 + 5.88\times10^4 \approx 1.6 \times 10^5$ Pa.)
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