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lubasha [3.4K]
4 years ago
15

Describe the relationship between work and power.

Physics
1 answer:
kakasveta [241]4 years ago
7 0
Work<span> done is generally referred in </span>relation<span> to the force applied while energy is used in reference to other factors such as heat. </span>Power<span> is defined as </span>work<span> done per unit time. Or energy per unit time. Eg: if you supply 50 joules over a period of 5 seconds, you have supplied 50/5=10 Watts of </span>power<span>.</span>
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Which type of wave is described as a wave that is not confined to the space along medium?
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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
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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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