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3 years ago

7

The legend that Benjamin Franklin flew a kite as a storm approached is only a legend; he was neither stupid nor suicidal. Suppos

e a kite string of radius 2.15 mm extends directly upward by 0.831 km and is coated with a 0.519 mm layer of water having resistivity 183 Ω m. If the potential difference between the two ends of the string is 166 MV, what is the current through the water layer?

1 answer:

UNO [17]3 years ago

3 0

Answer:

The current on the water layer = 1.64×10^-3A

Explanation:

Let's assume that the radius given for the string originates from the centre of the string. The equation for determining the current in the water layer is given by:

I = V × pi[(Rwater + Rstring)^2 - (Rstring)^2/ ( Resitivity × L)

I =[ 166×10^6 ×3.142[(0.519×10^-4) + (2.15×10^-3])^2 - ( 2.15×10^-3)^2] / ( 183 × 831)

I =[ 521572000(4.848×10^6)- 4.623×10^-6]/ 154566

I = 252.83 -(4.623×10^-6)/ 154566

I = 252.83/154566

I = 1.64× 10^-3A

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Read 2 more answers

A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o

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$\delta = \frac{PL}{AE}$

Where,

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Our values are given as,

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Replacing we have,

$\delta = \frac{PL}{AE}$

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$\delta = 0.001932m$

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