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Elden [556K]
3 years ago
7

Most of the resistance of the human body comes from the skin, as the interior of the body contains aqueous solutions that are go

od electrical conductors. For dry skin, the resistance between a person's hands is measured at typically 500 kΩ. The skin varies in thickness, but on the average it is about 2.0 mm thick. We can model the body between the hands as a cylinder 1.6 m long and 14 cmin diameter with the skin wrapped around it.
(a) What is the resistivity of the skin?
Ω · m

(b) Compare your answer with the internal resistivity of the body, approximately

4.8 Ω · m.

rhoskin
rhointerior
=

How do you explain the fact that the resistance of the skin is about 1000 times greater than the internal resistance of the body, even though the resistivity of the skin is only about 60 times that of the interior of the body?

A )The skin has a smaller cross sectional area.

B )The resistance of an object scales exponentially with respect to the resistivity of the material out of which it is made.

C ) The skin is constantly moistened by biological processes.

D ) Charges can only exist on the surface of conductors.
Physics
1 answer:
ruslelena [56]3 years ago
4 0

Answer:

a) Resistivity=R.A/L

Here, R=500000ohm, L=1.6m, A=2pi* (0.14/2)*0.002 m2=0.00088.........we have converted quantities into SI units

Resistivity=R.A/L

=500000*0.00088/1.6 = 440/1.6= 275 ohm.meter

b) pskin/pinterior =275/4.8=57.3

For second part, correct option is (A), because cross section area is smaller.

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A traveling electromagnetic wave in a vacuum has an electric field amplitude of 93.3 V/m. Calculate the intensity S of this wave
7nadin3 [17]

Answer:

Intensity = 11.56W/m²

The energy flowing through the given area is 4.55 J

Explanation:

The expression for the intensity of the electromagnetic wave is,

I = \frac{1}{2} C{ {\varepsilon _0}E_m^2

Here,\varepsilon _0 is the permittivity of the free space,

E_m  is the electric field amplitude and

c is the speed of the light.

substitute

⁸m/s for c

8.85×10  −12  C² /N⋅m² for {\varepsilon _0}

and 93.3 V/m for {E_{\rm{m}

I = \frac{1}{2} \times (3\times10^8)\times(8.85\times10^-^1^2)(93.3)\\\\I = 11.56W/m^2

The expression for the energy is,

E = I×A×t

Here, I is the intensity of the electromagnetic wave,

A is the area, and

t is the time.

Substitute

11.56W/m² for I

0.0287m ² for A

13.7s for t

E = (11.56)\times(0,0287)\times(13.7)\\E = 4.55J

The energy flowing through the given area is 4.55 J

5 0
4 years ago
If the earth’s tilt on it axis were to increase by 20 degrees, what would happen to earth?
Triss [41]

Each pole would become tropical during summer, but be horribly cold during winter, far colder than they are now. But the equator would be in perpetual twilight during two seasons—and summer the opposing two.

4 0
3 years ago
A ski gondola is connected to the top of a hill by a steel cable oflength 620 m and diameter 1.5 cm. As thegondola comes to the
klemol [59]

Answer:

Explanation:

Given

Length of cable L=620 m

Diameter of cable d=1.5 cm

time taken to return to original position T=14 s

time taken to cover distance L

t=\frac{T}{2}=7 s

velocity

v=\frac{L}{t}=\frac{620}{7}=88.57 m/s

(b)Relation between velocity of wave Tension is

v=\sqrt{\frac{T}{\mu }} , where \mu =mass per unit Length

T=v^2\cdot \mu

T=(88.57)^2\cdot \frac{m}{L}

T=(88.57)^2\cdot \frac{\rho AL}{L}

where \rho =density\ of\ steel =7850 kg/m^3

A=area\ of\ cross-section=\frac{\pi }{4}d^2=1.76\times 10^{-4} cm^2

T=(88.57)^2\cdot 7850\times 1.76\times 10^{-4}

T=10,883 N

5 0
3 years ago
Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration
stepan [7]

Answer:

The magnitude of the acceleration ae of the earth due to the gravitational pull of the moon is \mathbf{3.3187\times10^{-5}}\frac{m}{s^{2}}

Explanation:

By Newton's gravitational law, the magnitude of the gravitational force between two objects is:

F=G\frac{Mm}{r^{2}}(1)

With G the gravitational constant, M the mass of earth, m the mass of the moon and r the distance between the moon and the earth, a quick search on physics books or internet websites give us the values:

M=5.972\times10^{24}\,kg

m=7.34767309\times10^{22}\,kg

r=384400\,km

G=6.674\times10^{-11}\,\frac{N\,m^{3}}{kg^{2}}

Using those values on (1)

F=(6.674\times10^{-11})*\frac{(5.972\times10^{24})(7.34767309\times10^{22})}{(384400\times10^{3})^{2}}

F\approx1.98193\times10^{20}N

Now, by Newton's second Law we can find the acceleration of earth ae due moon's pull:

F=M*ae\Longrightarrow ae=\frac{F}{M}=\frac{1.98193\times10^{20}}{5.972\times10^{24}}\approx\mathbf{3.3187\times10^{-5}}

6 0
4 years ago
Can you guys help me with this guestion?
OlgaM077 [116]

Answer:

Explanation:

A -heat capacity

B - continuity equation

C -pressure

D - mechanical energy

3 0
3 years ago
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