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The typical design burst pressure for a rupture disc is what?

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Final answer:

The design burst pressure for a rupture disc is the pressure at which it is intended to burst. For a 5000-N force and a disk with a 2 cm radius, the pressure is approximately 3,978,875 Pa and the deformation is about 2.12 mm when considering a disc that is 0.800 cm thick with a Young's modulus of 1.5 x 10¹ N/m².

Step-by-step explanation:

The typical design burst pressure for a rupture disc is a prespecified pressure at which the disc is designed to fail to prevent pressure buildup and potential damage or explosion. In the context of the question provided, where a disk between spinal vertebrae is subjected to a 5000-N compressional force with a uniform circular cross section of 2.00 cm in radius:
(a) The pressure can be calculated using the formula P = F/A, where P is pressure, F is the force, and A is the area of the cross-section. Here, the area A is πr², where r is the radius. The pressure created can be calculated as:

P = 5000 N / (π x (0.02 m)²) = 5000 N / (π x 0.0004 m²) = 5000 N / (0.00125664 m²) = 3978874.64 Pa

(b) The deformation (ΔL) caused by the force can be calculated using the formula: ΔL = (F/A) * (L/Y), where L is the original thickness of the disk and Y is Young's modulus. Thus, we have:

ΔL = (5000 N / (0.00125664 m²)) * (0.008 m / (1.5 x 10¹ N/m²)) = 3978874.64 Pa * (0.008 m / 1.5 x 10¹ Pa) = 0.0212 m or 2.12 mm

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