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solong [7]
3 years ago
9

Determine the ratio of the resistivity of pure water to silver?

Physics
1 answer:
shutvik [7]3 years ago
3 0
Silver is a very good conductor, this means its resistivity is very low (from table, we can check the precise value, which is \rho_s = 1.6 \cdot 10^{-8} \Omega m).

Pure water, instead, is a very bad conductor, this means its resistivity is very high, of order of k \Omega \cdot m (10^3 \Omega m ). Even without knowing the precise value of the pure water resistivity, we can estimate the ratio between the pure water resistivity and the silver resistivity by comparing the two orders of magnitude:
r= \frac{10^3 \Omega m}{10^{-8} \Omega m}  \sim 10^{11}

Therefore, we can say that the correct answer is
3 \cdot 10^{11} : 1
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The 480 g bar is rotating as shown what is the angular momentum of the bar about the axle?
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On a similar problem wherein instead of 480 g, a 650 gram of bar is used:

Angular momentum L = Iω, where 
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
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