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Anestetic [448]
3 years ago
9

A current of 0.001 A can be felt by the human body. 0.005 A can produce a pain response. 0.015 A can cause a loss of muscle cont

rol. In the procedures of this lesson, over 0.030 A of current traveled in the three-battery circuit. Why was this circuit safe to handle with dry hands? Include the resistance of human skin in your answer.
Physics
1 answer:
lys-0071 [83]3 years ago
7 0

Answer:

In summary, it is safe to handle this voltage with dry hands because the current value that you pass through the body is smaller than its underestimated sensitivity.

Explanation:

The current flowing through your system is described by Ohm's law

        V = I R

where I is the current, V the voltage and R the resistance

in this case three barateras are taken in series giving a total voltage of V = 4.5 V the typical resistance values ​​of dry skin is R = 1000 000Ohm and the resinification of wet skin is R = 100000 ohm

let's calculate the current flowing

        I = V / R

        I = 4.5 / 1000000

        I = 4.5 10⁻⁶ A

this is the current with dry hands, we see that much less than the value that allows to feel a painful response by the body

If the skin is

         I = 4,5 / 100,000

         I = 4.5 10⁻⁵ A

This value is small, but it is close to the pain threshold, but it is in the range of slight discomfort.

In summary, it is safe to handle this voltage with dry hands because the current value that you pass through the body is smaller than its underestimated sensitivity.

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3 years ago
A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
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Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

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c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

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d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

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     Emo = E_{mf}

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     Y = y- 0.80

     Y = 11.8367 -0.80

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Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

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8 0
3 years ago
block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8
dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: \sqrt{800} \,\,\,N

Explanation:

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x-component of F1 is    F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

y-component of F1 is    F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

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y-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

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x-component of the velocity is:     v_x=4\,*\,2=8\,m/s

y-component of the velocity is:     v_y=4\,*\,2=8\,m/s

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