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Humans can bite with a force of approximately 800 N. If a human tooth has the Young's modulus of bone, a cross-sectional area of 1.0 square cm, and is 2.0 cm long, determine the change in the tooth's length during an 8.0 × 10^2 N bite.

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Final answer:

To find the change in a human tooth’s length under an 800 N bite, you can use the formula for elastic deformation, resulting in a change of approximately 10.67 μm.

Step-by-step explanation:

The question asks for the determination of the change in a human tooth’s length given a force of 800 N during a bite, the Young's modulus of bone, the cross-sectional area of the tooth, and its length. To find the change in length, we use the formula ΔL = FL/(AE) where ΔL is the change in length, F is the force applied, L is the original length, A is the cross-sectional area, and E is Young's modulus. Assuming that the Young's modulus for bone is about 1.5 × 1010 N/m2.

First, we convert the cross-sectional area from cm2 to m2 by multiplying with (1 m/100 cm)2. With the given values:

  • A = 1.0 cm2 = 1.0 × 10-4 m2
  • L = 2.0 cm = 2.0 × 10-2 m
  • F = 800 N
  • E = 1.5 × 1010 N/m2

Thus, ΔL = (800 N)(2.0 × 10-2 m) / (1.0 × 10-4 m2)(1.5 × 1010 N/m2) = 1.067 × 10-5 m or 10.67 μm.

The change in the tooth's length during an 800 N bite is approximately 10.67 micrometers (μm).

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