1. A patient's leg was put into traction, stretching the femur from a

1.	1. A patient's leg was put into traction, stretching the femur from a length of 0.46 m to 0461 m.The femur has a diameter of 3.05 cm. With the knowledge that bone has a Young's modulus of1.6 10 in tensions, what force was used to stretch the femur?
Answer» Thus the AMOUNT of force used to stretch the FEMUR is F = 25, 391 N Explanation: F = Y ( Δ L / LO) A Y = 1.6 x 10^10 Pa Lo = 0.46 m Δ L = Lo - L = (0.461 - 0.46 ) = 0.00 1 m A = πr^2 A = 3.14 x (1.525)^2 A = 7.30 cm^2 = 0.00073 m^2 Now using Young's modulus equation. F = 1.6 x 10^10 x (0.001 / 0.46 ) x 0.00073 F = 25, 391 N Thus the amount of force used to stretch the femur is F = 25, 391 N

1. A patient's leg was put into traction, stretching the femur from a length of 0.46 m to 0461 m.The femur has a diameter of 3.05 cm. With the knowledge that bone has a Young's modulus of1.6 10 in tensions, what force was used to stretch the femur?

Answer»

Thus the AMOUNT of force used to stretch the FEMUR is F = 25, 391 N

Explanation:

F = Y ( Δ L / LO) A

Y = 1.6 x 10^10 Pa

Lo = 0.46 m

Δ L = Lo - L = (0.461 - 0.46 ) = 0.00 1 m

A = πr^2

A = 3.14 x (1.525)^2

A = 7.30 cm^2 = 0.00073 m^2

Now using Young's modulus equation.

F = 1.6 x 10^10 x (0.001 / 0.46 ) x 0.00073

F = 25, 391 N

Thus the amount of force used to stretch the femur is F = 25, 391 N

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1. A patient's leg was put into traction, stretching the femur from a length of 0.46 m to 0461 m.The femur has a diameter of 3.05 cm. With the knowledge that bone has a Young's modulus of1.6 10 in tensions, what force was used to stretch the femur?​

Thus the AMOUNT of force used to stretch the FEMUR is F = 25, 391 N

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1. A patient's leg was put into traction, stretching the femur from a length of 0.46 m to 0461 m.The femur has a diameter of 3.05 cm. With the knowledge that bone has a Young's modulus of1.6 10 in tensions, what force was used to stretch the femur?