1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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.

You might be interested in
To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?
Nimfa-mama [501]
The energy stored in a capacitor is given by:
U= \frac{1}{2}CV^2
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
C=3.0 \mu F = 3.0 \cdot 10^{-6} F
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V
8 0
3 years ago
About how much time elapses between high tides on the same day?
Pani-rosa [81]
There is about 12 hours
6 0
3 years ago
A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0
3 years ago
Jenise is buying a car for $7,020. The TAVT rate is 9.1%.
djyliett [7]

Answer:

$7,658.82

Explanation:

<u>Sales Tax Calculations:</u>

Sales Tax Amount = Net Price x (Sales Tax Percentage / 100)

Total Price = Net Price + Sales Tax Amount

Net Price: $ 7,020.00

+Sales Tax (9.1%): $ 638.82

Total Price: $ 7,658.82

Therefore, the amount of tax that Jenise has to pay on her car is $7,658.82

8 0
3 years ago
If you throw a ball down ward then acceleration immeditely after leaving your hand is
bezimeni [28]

Answer:

9.8m/s²

Explanation:

The acceleration of the ball thrown after leaving my hand is 9.8m/s². This will be the acceleration due to gravity on the body.

  • Acceleration due to gravity is caused by the pull of the earth on a massive object.
  • The value of this acceleration is 9.8m/s².
  • As the ball nears the surface, it comes near zero.
7 0
3 years ago
Other questions:
  • What did the scientist Robert Hooke claim ?
    6·2 answers
  • Which of the following describes a condition in which an individual would not hear an echo?
    12·2 answers
  • A 10 kg block is pushed with a constant horizontal force of 20 n against a constant horizontal frictional force of 10 n. What is
    9·1 answer
  • What is difference between static electricity and current elecity...???​
    11·2 answers
  • 1. The maximum running speed (S) in km/hr and the corresponding body mass (m) in kg were
    6·1 answer
  • An insulated container with a divider in the middle contains two separated gases. Gas 1 is initially at a higher temperature tha
    15·2 answers
  • Which class of lever do spoon and scissors belong to?​
    5·2 answers
  • Plz help, and show work
    15·1 answer
  • A wave has frequency of 50 Hz and a wavelength of 10 m. What is the speed of the wave? Group of answer choices
    10·1 answer
  • PLEASE ANSWER NEED HELP!!!!!!!! PLEASE THE CORRECT ANSWER!!!!!!
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!