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AveGali [126]
3 years ago
9

Electric current passing through a human body can be dangerous, even fatal, depending on the amount of current, the duration of

the current, and the region of the body through which the current passes. However, currents less than about 0.5 mA are typically imperceptible. A current caused by a low potential difference typically travels through the outer layer of the skin. The resistance of the skin therefore determines how much current flows for a given potential difference. The resistance can be influenced by a number of factors, including whether the skin is wet or dry. For example, suppose an electric current of 89.0 µA follows a path through the thumb and index finger. When the skin is dry, the resistance along this path is 4.70 ✕ 105 Ω. What voltage (in V) is required for this current, in the case of dry skin? V When the skin is wet, the resistance is lowered to 2,200 Ω. What voltage (in V) is required for the same current, in the case of wet skin? V
Physics
1 answer:
Sati [7]3 years ago
7 0

Answer:

V for dry skin = 41.83 volts

V for wet skin = 0.196 volts

Explanation:

The relationship between current and voltage is given as follows by Ohm's Law:

V = IR

where,

V = voltage

I = Current

R = Resistance

FOR DRY SKIN:

V = Vd = ?

I = 89 μA = 89 x 10⁻⁶ A

R = 4.7 x 10⁵ Ω

Therefore,

Vd = (89 x 10⁻⁶ A)(4.7 x 10⁵ Ω)

<u>Vd = 41.83 volts</u>

FOR WET SKIN:

V = Vw = ?

I = 89 μA = 89 x 10⁻⁶ A

R = 2200 Ω

Therefore,

Vd = (89 x 10⁻⁶ A)(2200 Ω)

<u>Vd = 0.196 volts</u>

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¹/₃₈₇ second

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So, simply, the frequency is ¹/₃₈₇ second(s), as that is the reciprocal of the frequency, 387 Hz.

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Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on
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Answer:

8.8\times 10^{-6} W/m^2

Explanation:

We are given that

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1 m=100 cm

d=0.640 mm=0.64\times 10^{-3} m

1 mm=10^{-3} m

a=0.434 mm=0.434\times 10^{-3} m

y=0.830 mm=0.830\times 10^{-3} m

I_0=5.00\times 10^{-4}W/m^2

tan\theta=\frac{y}{R}

\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}

\theta=1.1\times 10^{-3}rad

tan\theta\approx \theta

Because \theta is small.

\phi=\frac{2\pi dsin\theta}{\lambda}

\sin\theta\approx \theta,

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\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}

\phi=7.74 rad

\beta=\frac{2\pi a\theta}{\lambda}

\beta=5.3 rad

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I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2

I=8.8\times 10^{-6} W/m^2

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B. Newton's First Law, I'm pretty sure. The first states that an object in motion stays in motion, and an object at rest stays at rest until an outside force is applied, and that seems pretty relevant.
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Answer:

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Explanation:

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Answer:

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