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Cerrena [4.2K]
3 years ago
10

The resistance of physiological tissues is quite variable. The resistance of the internal tissues of humans, primarily composed

of salty solutions, is quite low. Here the resistance between two internal points in the body is on the order of 100 ohms. Dry skin, however, can have a very high resistance, with values ranging from thousands to hundreds of thousands of ohms. However, if skin is wet, it is far more conductive, and so even contact with small voltages can create large, dangerous currents though a human body. (For example, although there is no specific minimum current that is lethal, currents generally exceeding a couple tenths of Amps may be deadly.)
Assuming that electrocution can be prevented if currents are kept below 0.1 A, and assuming the resistance of dry skin is 100,000 ohms, what is the maximum voltage with which a person could come into contact while avoiding electrocution? (Of course, all bets are off and things become far more dangerous if this person's skin is wet, which can reduce the resistance by more than a factor of 100.)
Physics
1 answer:
irga5000 [103]3 years ago
4 0

Given Information:  

Current = I = 0.1 A

Resistance = R = 100 kΩ

Required Information:  

Voltage = V = ?

Answer:  

Voltage = V = 1000 V

Step-by-step explanation:  

We know that electrocution depends upon the amount of current flowing through the body and the voltage across the body.

V = IR

Where I is the current flowing through the body and R is the resistance of body.

If electrocution can be avoided when the current is below 0.1 A then

V = 0.1*10×10³

V = 1000 Volts

Therefore, 1000 V is the maximum voltage with which a person could come into contact while avoiding electrocution, any voltage more than 1000 V may result in fatal electrocution.

Also note that human body has very low resistance when the body is wet therefore, above calculated value would not be applicable in such case.

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How much tension must a rope withstand if it is used to accelerate a 6526 kg car vertically upwards at 8.9 m/s^2?
ExtremeBDS [4]

Answer:

The value is T =  122036.2 \  N

Explanation:

From the question we are told that

     The mass of the car is  m  = 6526 \  kg

      The acceleration  is  a=  8.9 \  m/s^2

Generally the net force applied on the rope is mathematically represented as

          F_{net}   =  T -  W

Here W is the weight of the car which is evaluated as

         W =  m * g

=>      W =  6526  * 9.8

=>       W =  63954.8 \  N

Generally the net force can also be mathematically represented as

       F =  m * a

So

        m * a  =  T  -  63954.8

=>     6526  *  8.9 =  T  -  63954.8

=>      T =  122036.2 \  N

7 0
3 years ago
Professional application dr. john paul stapp was u.s. air force officer who studied the effects of extreme deceleration on the h
aleksley [76]
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Answer:

mas of water displaced = 41.4 g

Explanation:

Weight in air = True weight = 45 g

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8 0
3 years ago
Suppose the wavelength of the light is 460 nm. how much farther is it from the dot on the screen in the center of fringe e to th
gavmur [86]

The difference is 920nm.

<h3>What is called Wavelength?</h3>

The wavelength can be defined as the separation between wave crests. Additionally, a wide variety of objects move in waves, including light, the earth or ground, water, strings, air, and sound waves. Furthermore, we use the Greek letter lambda to denote the wave's wavelength. In addition, the wave's wavelength is determined by dividing its velocity by its frequency. Additionally, we employ meters (m), which is the unit used to indicate wavelength in meters.

Positive interference is produced by a bright fringe, and the wavelength is always multiple.

The difference at the center fringe is (460 nm)(0) = 0.

The difference for the first maximum is (460 nm)(1) = 460 nm.

The difference for the second maximum is (460 nm)(2) = 920 nm.

As a result, the difference for maxima n is (460 nm) (n)

Since C is on the second maximum in the image that was submitted, the difference is stated as (460 nm)(2) = 920 nm.

Consequently, the screen's central dot on the fringe E to the left slit is 920 nm.

To learn more about Wavelength, visit:

brainly.com/question/13533093

#SPJ4

3 0
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