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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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Explanation:

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3 years ago
A kayaker moves 22 meters northward, then 18 meters southward , and finally 24 meters northward what is the distance and magnitu
Elina [12.6K]

Answer:

Distance = 64 metres

Displacement = 28 metres

Direction = northward

Explanation:

Given that a  kayaker moves 22 meters northward, then 18 meters southward , and finally 24 meters northward

Distance covered is only about the magnitude of the length covered.

Distance = 22 + 18 + 24 = 64 metres

The direction will be considered when calculating the displacement

Let northward be positive and southward be negative

Displacement = 22 - 18 + 24 = 28 metres

Since the displacement is positive, the direction of the motion is northward.

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3 years ago
One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction
goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

0=F-FrictionAB

F=FrictionAB=Nab*μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

0=Nab-W

Nab=W=mg

replacing this we have:

F=μs*Nab=μs*mg=41.6N

and  μs=41.6N/(mg)

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

F-Friction1(ground)-Friction2(AB)=0

This is the new external F force that we are looking for:

F=Friction1(ground)+Friction2(AB)

we know Friction2(AB) because we found that in the previous block so:

F=Friction1(ground)+mg*μs

for the other friction we have to use the equation:

Friction(ground)=N(ground)*μs

from y axis we have:

N(ground)-w-Normal(AB)=0

N(ground)=w+Normal(AB)

we found the value of Normal(AB) with the previous block so:

N(ground)=mg+mg=2mg

and:

Friction(ground)=2mg*μs

F=Friction(ground)+mg*μs

F=2mg*μs+μs*mg=3mg*μs

and since: μs*mg=41.6N

the new F force would be:

F=3mg*μs=41.6*3=124.8N

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Answer it asap<br> I promise i will mark them the Brainly
nataly862011 [7]

Answer:

A) The acceleration is zero

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Explanation:

<u>Velocity vs Time Graph</u>

It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.

The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.

A. To calculate the acceleration, we use the formula:

\displaystyle a=\frac{v_2-v_1}{t_2-t_1}

Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:

\displaystyle a=\frac{8-8}{12-0}=0

This confirms the previous conclusion.

B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.

Area of rectangle= base*height=12 s * 8 m/s = 96 m

Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m

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A cyclist travels at 15 m/s during a sprint finish. What is this speed in km/h
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Answer:

54 km/hr

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15 m/s to km/hr => 15 x 18/5 =>3 x 18 => 54km/hr

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