The Achilles tendon is a fibrous tissue that connects your soleus to your calcaneus and can withstand 145 MPa of tensile stress. Assume the Achilles tendon can be modeled as a cylinder with a diameter with a diameter of 3.25 cm. How much normal tensile force (in Newtons) would the soleus need to generate to tear the Achilles tendon

Answers

Answer 1

Answer:

 F = 1.2 10⁵ N

Explanation:

Force is defined by

          P = F / A

          F = P A

The area is

          A = π r²

the radius is half the diameter

          r = d / 2

          A = π d²/4

         

we substitute

          F = P π d²/4

let's calculate

          F = 145 10⁶  π (3.25 10⁻²)²/4

          F = 1.2 10⁵ N


Related Questions

The skater lowers her arms as shown in the adjacent
figure decreasing her radius to 0.15 m. Find her new speed.

Answers

Answer:

is there more?

Explanation:

7. Consumers can be predators, prey, scavengers, herbivores omnivoree, or carnivore: What is the common characteristic of all consumer: A Create their own food B. Must eat to get energy C. Hunt for live animals D. Will eat anything​

Answers

Answer:

B. Must eat to get energy

Explanation:

The common characteristics of all consumers is that they must eat to get energy. This way, they are termed heterotrophs.

Heterotrophs are organisms that cannot make their own food. They must eat other organisms to obtain nutrition for energy needs.

Plants do not do this. They are autotrophs in that they simply make their own food.

They use this food to obtain energy for their living activities.

How does the size of a wind turbine affect its energy output?
A.)Smaller turbines spin slower.
B.) Larger turbines have a greater storage capacity.
C.) Larger turbines generate more electricity.
D.)Smaller turbines are better for capturing strong winds.

Answers

Answer:

Larger tubines generate more electricity.

Explanation:

Larger blades allow the turbine to capture more of the kinetic energy of the wind by moving more air through the rotors. However, larger blades require more space and higher wind speeds to operate. This distance is necessary to avoid interference between turbines, which decreases the power output.

The average marathon runner can complete the 42.2-km distance of the marathon in 3 h and 30 min. If the runner's mass is 85 kg, what is the runner's average kinetic energy during the run

Answers

Answer:

the runner's average kinetic energy during the run is 476.96 J.

Explanation:

Given;

mass of the runner, m = 85 kg

distance covered by the runner, d = 42.2 km = 42,200 m

time to complete the race, t = 3 hours 30 mins = (3 x 3600s) + (30 x 60s)

                                                                               = 12,600 s

The speed of the runner, v = d/t

                                          v = 42,200 / 12,600

                                          v = 3.35 m/s

The runner's average kinetic energy during the run is calculated as;

K.E = ¹/₂mv²

K.E = ¹/₂ × 85 × (3.35)²

K.E = 476.96 J

Therefore, the runner's average kinetic energy during the run is 476.96 J.

which form of energy is an example of kinetic energy

Answers

Answer:

1. realizing of arrow

2. kicking of ball

3. punching the punching bag

On a sunny day, a rooftop solar panel delivers 60 W of power to the house at an emf of 17 V. How much current flows through the panel

Answers

Answer:

3.53 amps

Explanation:

Given data

Power= 60W

Voltage= 17V

The expression relating current, power, and voltage is

P= IV

substitute

60= I*17

I= 60/17

I= 3.53 amps

Hence the current that flows is 3.53 amps

Two coils have the same number of circular turns and carry the same current. Each rotates in a magnetic field acting perpendicularly to its axis of rotation. Coil 1 has a radius of 4.5 cm and rotates in a 0.21-T field. Coil 2 rotates in a 0.39-T field. Each coil experiences the same maximum torque. What is the radius (in cm) of coil 2

Answers

Answer:

Explanation:

Torque acting on a coil in a magnetic field = MBsinθ where M is magnetic moment , B is magnetic field and θ is inclination of the normal to coil with direction of field.

For maximum torque sinθ = 1

Maximum torque = MB

M = NIA where N is no of turns , I is current and A is area of the coil

Maximum torque = NIAB

As maximum torque is same

N₁I₁A₁B₁ = N₂I₂A₂B₂

N₁ = N₂ , I₁ = I₂

A₁B₁ = A₂B₂

π R₁² B₁ = π R₂² B₂

4.5² x .21 = R₂² x .39

R₂² = 10.9

R₂ = 3.3 cm .

The water side of the wall of a 60-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the dam and its line of action when the dam is filled to the rim. Take the density of water to be 1000 kg/m3.

Answers

Answer:

[tex]26852726.19\ \text{N}[/tex]

[tex]57.52^{\circ}[/tex]

Explanation:

r = Radius of circle = 7 m

w = Width of dam = 60 m

h = Height of the dam will be half the radius = [tex]\dfrac{r}{2}[/tex]

A = Area = [tex]rw[/tex]

V = Volume = [tex]w\dfrac{\pi r^2}{4}[/tex]

Horizontal force is given by

[tex]F_x=\rho ghA\\\Rightarrow F_x=1000\times 9.81\times \dfrac{7}{2}\times  7\times 60\\\Rightarrow F_x=14420700\ \text{N}[/tex]

Vertical force is given by

[tex]F_y=\rho gV\\\Rightarrow F_y=1000\times 9.81\times 60\times \dfrac{\pi 7^2}{4}\\\Rightarrow F_y=22651982.59\ \text{N}[/tex]

Resultant force is

[tex]F=\sqrt{F_x^2+F_y^2}\\\Rightarrow F=\sqrt{14420700^2+22651982.59^2}\\\Rightarrow F=26852726.19\ \text{N}[/tex]

The hydrostatic force on the dam is [tex]26852726.19\ \text{N}[/tex].

The direction is given by

[tex]\theta=\tan^{-1}\dfrac{F_y}{F_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{22651982.59}{14420700}\\\Rightarrow \theta=57.52^{\circ}[/tex]

The line of action is [tex]57.52^{\circ}[/tex].

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