Question 28 1 pts A planet has an atmospheric pressure of $50,000 \mathrm{~Pa}$, and an acceleration due to gravity half that on the earth's surface. What is the longest drinking straw that could be used to suck up a liquid of density $500 \mathrm{~kg} / \mathrm{m}^{3}$ ? $\mathrm{m}$

Step 1 ：The atmospheric pressure (P) on the planet is given as 50,000 Pa.

Step 2 ：The density (ρ) of the liquid is given as 500 kg/m³.

Step 3 ：The acceleration due to gravity (g) on the planet is half that on Earth, which is 9.8 m/s². Therefore, g = 9.8 / 2 = 4.9 m/s².

Step 4 ：The maximum height of a column of liquid that can be supported by atmospheric pressure is given by the formula: \(h = \frac{P}{ρg}\).

Step 5 ：Substituting the given values into the formula, we get \(h = \frac{50000}{500 \times 4.9}\).

Step 6 ：Solving the equation gives us the maximum height of the liquid column, which is approximately 20.41 m.

Step 7 ：Final Answer: The longest drinking straw that could be used to suck up a liquid of density 500 kg/m³ on this planet is \(\boxed{20.41 \mathrm{m}}\).