Determine the effective modulus of elasticity of the sartorius muscles

It is necessary to determine the effective modulus of elasticity of the frog sartorius muscle. To do this, consider the data: when the tension on the muscle increases from 10 kPa to 40 kPa, its length increases from 0.032 m to 0.034 m.

To calculate the effective modulus of elasticity, we use the formula:

E = (F*L)/(S*ΔL)

Where:

• E - effective modulus of elasticity;
• F - applied force;
• L - initial length of the muscle;
• S - cross-sectional area of ​​the muscle;
• ΔL - change in muscle length.

According to the condition, the initial length of the muscle is known (L = 0.032 m), change in muscle length (ΔL = 0.034 m - 0.032 m = 0.002 m) and applied voltage (F/S = 10 kPa = 10⁴ N/m²). It is necessary to find the cross-sectional area of ​​the muscle (S) and effective modulus of elasticity (E).

From the formula for the deformation of a solid body ε = ΔL/L let's find the lengthening of the muscle:

ε = ΔL/L = 0,002 м/0,032 м = 0,0625

From Hooke's formula for the modulus of elasticity E = s/e Let's find the effective modulus of elasticity:

E = σ/ε = (F/S)/ε = (10⁴ N/m²)/(0,0625) = 1,6 * 10⁵ N/m² = 160 kPa

Thus, the effective modulus of elasticity of the frog sartorius muscle is 160 kPa.

Determine the effective modulus of elasticity of the sartorius muscles

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To determine the effective modulus of elasticity of the frog sartorius muscle, you must use the formula:

E = (FL)/(SΔL)

Where E is the effective modulus of elasticity, F is the applied force, L is the initial length of the muscle, S is the cross-sectional area of ​​the muscle, ΔL is the change in muscle length.

From the problem conditions it is known that the initial muscle length L = 0.032 m, the change in muscle length ΔL = 0.034 m - 0.032 m = 0.002 m, and the applied voltage F/S = 10 kPa = 104 N/m^2.

To determine the cross-sectional area of ​​the muscle, it is necessary to express it from the applied voltage formula:

F/S = (P/S)(S(ΔL/L))

Where P is the perimeter of the cross section of the muscle. Thus, the cross-sectional area of ​​the muscle is:

S = (P*ΔL)/(F/L)

Since the cross-section of a frog muscle is approximately circular, its perimeter can be estimated using the formula for circumference:

P = π*D

Where D is the cross-sectional diameter of the muscle. Approximately, we can assume that the diameter is equal to the initial length of the muscle, i.e. D = L.

Substituting known values ​​into the formulas, we get:

P = πL = 0,1005 м S = (PΔL)/(F/L) = (0.1005 m * 0.002 m)/(104 N/m^2 / 0.032 m) ≈ 0.00019 m^2 ε = ΔL/L = 0.002 m/0.032 m = 0.0625 E = σ/ε = (F/S)/ε = (104 N/m^2)/(0.0625) ≈ 1.6 * 10^5 N/m^2 = 160 kPa

Thus, the effective modulus of elasticity of the frog sartorius muscle is about 160 kPa.

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The modulus of elasticity of the frog sartorius muscle can be determined by the formula:

E = (F / A) / (ΔL / L)

where E is the elastic modulus, F is the force applied to the muscle, A is the cross-sectional area of ​​the muscle, ΔL is the change in muscle length, L is the initial length of the muscle.

From the problem it is known that with an increase in stress from 10 kPa to 40 kPa, the muscle increased in length from 0.032 m to 0.034 m. It is also necessary to know the cross-sectional area of ​​the muscle and the initial length, which are not given in the problem.

To solve the problem, it is necessary to use Hooke's law, which states that the deformation of a body is proportional to the force applied to it. You also need to know the formula for the area of ​​a circle:

A = πr^2,

where r - radius of the circle.

Thus, to solve the problem, it is necessary to know the cross-sectional area of ​​the muscle and its initial length in order to calculate the elastic modulus using the formula using Hooke’s law. If this data is missing, then the problem cannot be solved.

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