The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme a
ctivities, such as sprinting, it can be subjected to forces as high as 13.0 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is 78.2 mm^2 , its average length is 26 cm , and its average Young's modulus is 1474 MPa . How much tensile stress is required to stretch this muscle by 5.2% of its length? If we model the tendon as a spring, what is its force constant? If a 75 kg sprinter exerts a force of 13.0 times his weight on his Achilles tendon, by how much will it stretch?
1 answer:
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Answer:
a) 5.63 atm
Explanation:
We can use combined gas law
<em>The combined gas law</em> combines the three gas laws:
- Boyle's Law, (P₁V₁ =P₂V₂)
- Charles' Law (V₁/T₁ =V₂/T₂)
- Gay-Lussac's Law. (P₁/T₁ =P₂/T₂)
It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.
P₁V₁/T₁ =P₂V₂/T₂
where P = Pressure, T = Absolute temperature, V = Volume occupied
The volume of the system remains constant,
So, P₁/T₁ =P₂/T₂
a)
Answer:
- The maximum height reached by the ball is 45.92 m
- Time taken to fall down to half of its height is 2.2 s
Explanation:
Given;
initial velocity of the ball, u = 30 m/s
final velocity of the ball at the highest point, v = 0
The maximum height reached by the ball is calculated as;
v² = u² - 2gh
where;
h is the maximum height reached by the ball
0 = 30² - (2 x 9.8)h
19.6h = 900
h = 900 / 19.6
h = 45.92 m
Time taken to fall to half of its height is calculated as;
when falling down, the final velocity v becomes the initial velocity = 0.
Apply the following kinematic equation;
h = ut + ¹/₂gt²
h = 0 + ¹/₂gt²
h = ¹/₂gt²
where;
h = 45.92 m is the maximum height reached
half of h = 45.92 / 2 = 22.96 m
22.96 = ¹/₂gt²