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Vsevolod [243]
3 years ago
8

The conductive tissues of the upper leg can be modeled as a 40-cm-long, 12-cm-diameter cylinder of muscle and fat. The resistivi

ties of muscle and fat are 13 Ω m and 25 Ω m, respectively. One person’s upper leg is 82% muscle, 18% fat. What current is measured if a 1.5 V potential difference is applied between the person’s hip and knee?
Physics
1 answer:
ahrayia [7]3 years ago
3 0

To calculate and solve the problem it is necessary to apply the concepts related to resistance and resistivity.

The equation that is responsible for relating the two variables is:

R = \rho \frac{L}{A}

Where,

R= Resistance of the conductor

\rho =Resistivity of the conductor material

L = Length

A = Cross-sectional area of conductor

With the previous values the area of the muscle (Real Muscle-82%)is,

A_m = (0.82)\pi r^2 = (0.82)\pi (12/2*10^{-2})^2

A_m = 9.274*10^{-3}m^2

Using the equation from Resistance we have that at the muscle the value is:

R_m = \rho \frac{L}{A}

R_m = \frac{13*(0.4)}{9.274*10^{-3}}

R_m = 560.70\Omega

At the same time we can make the same process to calculate the resistance of the fat, then

A_m = (0.18)\pi r^2 = (0.18)\pi (12/2*10^{-2})^2

A_m = 2.0357*10^{-3}m^2

The resistance of the fat would be,

R_f = \rho \frac{L}{A}

R_f = \frac{25*(0.4)}{2.0357*10^{-3}}

R_f = 4912.3\Omega

Then the total resistance in a set as the previously writen, i.e, in parallel is:

R=\frac{R_mR_f}{R_m+R_f}

R = \frac{(560.70)(4912.3)}{4912.3+560.70}

R = 502.62\Omega

We can here apply Ohm's law, then

I= \frac{V}{R}

I = \frac{1.5}{502.62}

I = 2.984*10^{-3}A

I = 2.984mA

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A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When
11111nata11111 [884]

Answer:

The value is R_f =  \frac{4}{5}  R

Explanation:

From the question we are told that

   The  initial velocity of the  proton is v_o

    At a distance R from the nucleus the velocity is  v_1 =  \frac{1}{2}  v_o

    The  velocity considered is  v_2 =  \frac{1}{4}  v_o

Generally considering from initial position to a position of  distance R  from the nucleus

 Generally from the law of energy conservation we have that  

       \Delta  K  =  \Delta P

Here \Delta K is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

      \Delta K  =  K__{R}} -  K_i

=>    \Delta K  =  \frac{1}{2}  *  m  *  v_1^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K  =  \frac{1}{2}  *  m  * (\frac{1}{2} * v_o )^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K  =  \frac{1}{2}  *  m  * \frac{1}{4} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2

And  \Delta  P is the change in electric potential energy  from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

          \Delta P =  P_f - P_i

Here  P_i is zero because the electric potential energy at the initial stage is  zero  so

             \Delta P =  k  *  \frac{q_1 * q_2 }{R}  - 0

So

           \frac{1}{2}  *  m  * \frac{1}{4} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2 =   k  *  \frac{q_1 * q_2 }{R}  - 0

=>        \frac{1}{2}  *  m  *v_0^2 [ \frac{1}{4} -1 ]  =   k  *  \frac{q_1 * q_2 }{R}

=>        - \frac{3}{8}  *  m  *v_0^2  =   k  *  \frac{q_1 * q_2 }{R} ---(1 )

Generally considering from initial position to a position of  distance R_f  from the nucleus

Here R_f represented the distance of the proton from the nucleus where the velocity is  \frac{1}{4} v_o

     Generally from the law of energy conservation we have that  

       \Delta  K_f  =  \Delta P_f

Here \Delta K is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus  , this is mathematically represented as

      \Delta K_f   =  K_f -  K_i

=>    \Delta K_f  =  \frac{1}{2}  *  m  *  v_2^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K_f  =  \frac{1}{2}  *  m  * (\frac{1}{4} * v_o )^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K_f  =  \frac{1}{2}  *  m  * \frac{1}{16} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2

And  \Delta  P is the change in electric potential energy  from initial position to a  position of  distance R_f  from the nucleus , this is mathematically represented as

          \Delta P_f  =  P_f - P_i

Here  P_i is zero because the electric potential energy at the initial stage is  zero  so

             \Delta P_f  =  k  *  \frac{q_1 * q_2 }{R_f }  - 0      

So

          \frac{1}{2}  *  m  * \frac{1}{8} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2 =   k  *  \frac{q_1 * q_2 }{R_f }

=>        \frac{1}{2}  *  m  *v_o^2 [-\frac{15}{16} ]  =   k  *  \frac{q_1 * q_2 }{R_f }

=>        - \frac{15}{32}  *  m  *v_o^2 =   k  *  \frac{q_1 * q_2 }{R_f } ---(2)

Divide equation 2  by equation 1

              \frac{- \frac{15}{32}  *  m  *v_o^2 }{- \frac{3}{8}  *  m  *v_0^2  } }   =  \frac{k  *  \frac{q_1 * q_2 }{R_f } }{k  *  \frac{q_1 * q_2 }{R } }}

=>           -\frac{15}{32 } *  -\frac{8}{3}   =  \frac{R}{R_f}

=>           \frac{5}{4}  =  \frac{R}{R_f}

=>             R_f =  \frac{4}{5}  R

   

7 0
3 years ago
A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.
SVETLANKA909090 [29]

A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

A)

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of x_0 = 5.0 m

- at t = 0, the initial velocity of the motorcyclist is v_0 = 15.0 m east

- The acceleration of the motorcyclist is constant and it is a=4.0 m/s^2

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

x(t)=x_0 + v_0t + \frac{1}{2}at^2

where t is the time.

Substituting t = 2.0 s, we find the position:

x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m

B)

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

v(t)=x'(t)=v_0 + at

where:

v_0=15.0 m/s is the initial velocity

a=4.0 m/s^2 is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s

And therefore, the position at t = 2.5 s is:

x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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A nebula is gas, dust, and other things that form and form some sort of clould like thing. It also looks cool because of the colors it produces. I'm sure you could do a web search to find more about it, I don't take astronomy hahaha
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We reduce friction in machines? why<br>​
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