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stellarik [79]
3 years ago
11

Tendons are, essentially, elastic cords stretched between two fixed ends; as such, they can support standing waves. These resona

nces can be undesirable. The Achilles tendon connects the heel with a muscle in the calf. A woman has a 20-cm long tendon with a cross-section area of 130 mm^2. The density of tendon tissue is 1100 kg/m^3.
Required:
For a reasonable tension of 600 , what will be the fundamental resonant frequency of her Achilles tendon?
Physics
1 answer:
Hatshy [7]3 years ago
3 0

Answer:

161.938 Hz

Explanation:

the computation of the fundamental resonant frequency is shown below

p = 1100 kg/m^3

A = 130 mm^2

= 130 ×10^-6 m^2

T = 600 N

L = 20 cm

= 0.2 m

Now the linear density of tendon is

=  1100 kg/m^3 ×  130 ×10^-6

= 0.143 kg/m

Now the wave of the string is

= √600  ÷ √0.143

= 64.775 m/s

Now finally the fundamental resonant frequency is

= 64.775 ÷ (2 × 0.2)

=161.938 Hz

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A government agency estimated that air bags have saved over 14,000 lives as of April 2004 in the United States. (They also state
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To solve this problem it is necessary to apply the concepts related to momentum, momentum and Force. Mathematically the Impulse can be described as

I = F*t

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F= Force

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P = m\Delta v \rightarrow P=m(v_1-v_2)

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From equilibrium the impulse is equal to the momentum, therefore

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PART A) Since the body ends at rest, we have the final speed is zero, so the momentum would be

p=m(v_1-v_2)

p = 75*0.15

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PART B) From the equation obtained previously we have that the Force would be:

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A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha
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Answer:

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  • The heat absorbed is 7475.69 joules

Explanation:

<h3>Work</h3>

We know that the differential work made by the gas  its defined as:

dW =  P \ dv

We can solve this by integration:

\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

P \ V = \ n \ R \ T

P = \frac{\ n \ R \ T}{V}

This give us

\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv

As n, R and T are constants

\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv

\Delta W= \ n \ R \ T  \left [ ln (V) \right ]^{v_2}_{v_1}

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ln (\frac{v_2}{v_1})

But the volume is:

V = \frac{\ n \ R \ T}{P}

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\Delta W = \ n \ R \ T  ln(\frac{P_1}{P_2})

Now, lets use the value from the problem.

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T = 27 \° C = 300.15 \ K

The ideal gas constant:

R = 8.314 \frac{m^3 \ Pa}{K \ mol}

So:

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E= c_v n R T

where c_v its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

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So:

\Delta E = 0

\Delta Q = \Delta W

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