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ollegr [7]
3 years ago
11

Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to

that between the large faces (D x L.) 2W. What is the ratio of the resistance as measured between the small faces (WxW) to that between the large faces (DxL)?
Physics
1 answer:
Ludmilka [50]3 years ago
6 0

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

        R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

         L = W

Area  

         A = W W = W²

        R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

       Length L = D= 2W

       Area     A = W L

     R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

    R₂ / R₁ = 2W²/L

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Consider velocity to the right as positive.

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Because momentum of the system is preserved, therefore
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Answer:
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3 years ago
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2 years ago
A runner drank a lot of water during a race. What is the expected path of the extra filtered water molecules?
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Explanation:

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Goormaghtigh cells are arranged at an angle between afferent and effector arterioles and meet in small columns. They are closely related to polar bearing cells. Between both formations is the dense macula (or Zimmerman's dense macula) that is in contact with the distal tubule and afferent arteriole just before it penetrates the glomerulus. These three formations, polar bearing, Goormaghtigh cells and dense macula form the juxtaglomerular apparatus that regulates the blood flow in the glomerulus.

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6 0
3 years ago
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Answer:

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