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Alla [95]
4 years ago
8

You have a set of calipers that can measure thicknesses of a few inches with an uncertainty of ± 0.005 inches. You measure the t

hickness of a deck of 52 cards and get 0.590 in:a. If you now calculate the thickness of 1 card, what is your answer, including its uncertainty?b. You can improve this result by measuring several decks together. If you want to know the thickness of 1 card with an uncertainty of only 0.00002 in., how many decks do you need to measure together?
Physics
1 answer:
Bond [772]4 years ago
3 0

Answer:

a) x = (0.0114 ± 0.0001) in , b) the number of decks is 5

Explanation:

a) The thickness of the deck of cards (d) is measured and the thickness of a card (x) is calculated

        x = d / 52

        x = 0.590 / 52

        x = 0.011346 in

Let's look for uncertainty

       Δx = dx /dd Δd

       Δx = 1/52 Δd

       Δx = 1/52  0.005

       Δx = 0.0001 in

The result of the calculation is

        x = (0.0114 ± 0.0001) in

b) You want to reduce the error to Δx = 0.00002, the number of cards to be measured is

           #_cards = n 52

The formula for thickness is

           x = d / n 52

Uncertainty

          Δx = 1 / n 52  Δd

         n = 1/52 Δd / Δx

         n = 1/52 0.005 / 0.00002

         n = 4.8

Since the number of decks must be an integer the number of decks is 5

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Given that:

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R = 1.29 cm

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Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

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The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

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V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

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Answer:

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Explanation:

The diagram for this question is shown on the first uploaded image

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From the question we are told that

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=> \frac{F_{T1}}{F_{T2}}  =  2

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