The plastic tube in the figure has a cross-sectional area of 5.59 cm2. The tube

Question

The plastic tube in the figure has a cross-sectional area of 5.59 cm2. The tube is filled with water until the short arm (of length d = 0.836 m) is full. Then the short arm is sealed and more water is gradually poured into the long arm. If the seal will pop off when the force on it exceeds 9.35 N, what total height of water in the long arm will put the seal on the verge of popping?




the tolerance is +/-2%

The plastic tube in the figure has a cross-sectional area of 5.59 cm2. The tube is filled with water until the short arm (of length d = 0.836 m) is full. Then the short arm is sealed and more water is gradually poured into the long arm. If the seal will pop off when the force on it exceeds 9.35 N, what total height of water in the long arm will put the seal on the verge of popping?




the tolerance is +/-2%

The plastic tube in the figure has a cross-sectional area of 5.59 cm2. The tube is filled with water until the short arm (of length d = 0.836 m) is full. Then the short arm is sealed and more water is gradually poured into the long arm. If the seal will pop off when the force on it exceeds 9.35 N, what total height of water in the long arm will put the seal on the verge of popping?




the tolerance is +/-2%

Explanation / Answer

Solution:

Area = A = 5.59 x 10-4 m2

distance = d = 0.36 m

Force = F = 9.35 N

Pressure = Force /Area

Volume of water that weighs 9.35 N =mass / density= [force / gravity] /density = [(9.35/9.8)] / 1000

V = 0.000954 m3 .

Area x height = Volume

height = h = volume / area = 0.000954 / (5.59x10^-4) = 1.71 m

Total height = H = 1.71 + 0.836 = 2.546 m

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