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Pavlova-9 [17]
3 years ago
10

A person notices a mild shock if the current along a path through the thumb and index finger exceeds 82 µA. Find the maximum all

owable potential difference without shock across the thumb and index finger for a dry-skin resistance of 2.4 × 105 Ω. Answer in units of V.
Physics
1 answer:
Drupady [299]3 years ago
4 0

Answer:

V_{voltage}=19.68V

Explanation:

Given data

Current I=82µA=82×10⁻⁶A

Resistance R=2.4×10⁵Ω

to find

Voltage

Solution

From Ohms law we know that:

V_{voltage}=I_{current}*R_{resistance}\\V_{voltage}=82*10^{-6}*2.4*10^{5}   \\V_{voltage}=19.68V

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David is investigating the properties of soil using the sample shown.
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<h3>What are these properties an example of?</h3>

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The atmosphere protects us from _____ and _____.
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7 0
3 years ago
Read 2 more answers
Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ
Tomtit [17]

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}

Put the value into the formula

525=\sqrt{\dfrac{c+v}{c-v}}\times950

\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2

\dfrac{c+v}{c-v}=0.305

c+v=0.305\times(c-v)

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v=\dfrac{0.305-1}{1+0.305}c

v=−0.532c

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

7 0
3 years ago
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