Question

A vertical U-shaped nibe is filled with a liquid of density p and the right end of the rube is sealed with a copper, as shown in the left figure. Some of the liquid is removed from the left column with a syringe and the left column descends a distance d, while the right column remains as it was. Refer to the right figure UE B 50% Part (a) Three students are debating Student A The pressure at level C must now be greater than the atmospheric pressure because liquid there is being pushed up against the stopper. Student B: I think the pressure at level B must be the same as at level A, because they are at the same level Both are at atmospheric pressure So the pressure at level C must be lower than atmospheric, because pressure decreases as you ascend Student C. But the liquid is denser than air, so the pressure at C cannot be less than atmospheric pressure. With which student(s) should you agree? B V Correct! D A 50%% Part (b) Calculate the pressure difference, in pascals, between levels C and A. AP - PC - PA. for p = 0.8/: 10' kgym and d - 4 cr Grade Summary

Answer 1

Student B is correct. The pressure at level B must be the same as at level A, both at atmospheric pressure. The pressure difference between levels C and A is approximately 0.03136 Pa.

Step-by-step explanation:

Student B is correct. The pressure at level B must be the same as at level A because they are at the same level, both at atmospheric pressure. Student A is incorrect in stating that the pressure at level C must be greater than atmospheric pressure. Since the liquid in the U-tube is denser than air, the pressure at level C cannot be less than atmospheric pressure. Therefore, you should agree with Student B.

To calculate the pressure difference between levels C and A, we can use the equation P = hpg, where P is the pressure difference, h is the height difference between the two levels, p is the density of the liquid, and g is the acceleration due to gravity. In this case, the height difference d is given as 4 cm, the density p is given as 0.8 kg/m³, and the acceleration due to gravity g is approximately 9.8 m/s². Plugging in these values, we can calculate the pressure difference:

P = (4/100) * 0.8 * 9.8 = 0.03136 Pa

Answer 2

The question discusses the pressures in a sealed U-shaped tube when the liquid level is lowered on one side. The pressure difference between levels C and A is calculated based on the density of the liquid and the distance the liquid dropped, resulting in 313.92 Pascals.

Step-by-step explanation:

The student is asking which student's argument is correct regarding the pressure changes in a U-shaped tube partially filled with liquid when one side is sealed with a stopper, and some liquid is removed from the other side. To answer part (b) of the question, we need to calculate the pressure difference between levels C and A in the U-shaped tube.

Given that the density (p) of the liquid is 0.8 x 103 kg/m^3, and the distance (d) the liquid column has fallen on the open side is 4 cm, we apply the formula for pressure difference (ΔP) in a liquid column which is ΔP = PC - PA = pgh, where g is the acceleration due to gravity (9.81 m/s^2) and h is the height difference between the two columns in meters (0.04 m).

To find ΔP, we multiply the density of the liquid by g and by h (0.8 x 103 kg/m^3 * 9.81 m/s^2 * 0.04 m), which gives a pressure difference of 313.92 Pascals (Pa).