Physics, 20.01.2020 20:31 thexgar

Ihave a set of calipers that can measure thickness of a few inches with an uncertainty of ±0.005 inches. i measure the thickness of a deck of 52 cards and get 0.590 inch. a) if i now calculate the thickness of 1 card, what is my answer (including uncertainty)b) i can improve this result by measuring several decks together. if i want to know the thickness of 1 card with an uncertainty of only 0.2 inch, how many decks do i need to measure?

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Answer from: arissahyp73x7x

The required number of decks to measure the uncertainty is 5.

Given data:

The number of cards is, 52.

Uncertainty with thickness is, 0.590 ±0.005 inches.

(a)

Since this is thickness of all the cards, to get the thickness of a single card we divide the total thickness by the number of cards. So uncertainty in thickness for 1 card is,

$u=\dfrac{0.590}{52} \pm \dfrac{0.005}{52}\\\\u=0.0113461 \pm 0.000096 \;\rm inches$

Considering that thickness is given to 3 significant figures while uncertainty is to 1 significant figure, the final answer should also confirm to these hence giving the first part of the answer to 3 significant figures while second part to 1 significant figure yields 0.0113 ±0.0001 inches.

Thus, we can conclude that the required thickness uncertainty of one card is 0.0113 ±0.0001 inches.

(b)

Let the cards have uncertainty of 0.0001 inches and the number of decks, n required to create uncertainty of 0.00002 inch is given as.

$n=0.0001/0.00002\\\\n=5$

Thus, we can conclude that the required number of decks to measure the uncertainty is 5.

Learn more about the uncertainties here:

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Answer from: Rogeartest4

(a) 0.0113 ±0.0001 inches

(b) 5 decks

Explanation:

Given information

52 cards, thickness is 0.590 ±0.005 inches

Since this is thickness of all the cards, to get the thickness of a single card we divide the total thickness (plus uncertainty) by the number of cards hence

1 card= $\frac {0.590}{52}$ ± $\frac {0.005}{52}$ = 0.0113461 ±0.000096 inches

Considering that thickness is given to 3 significant figures while uncertainty is to 1 significant figure, the final answer should also conform to these hence giving the first part of the answer to 3 significant figures while second part to 1 significant figure yields 0.0113 ±0.0001 inches

(b)

Considering that the cards have uncertainty of 0.0001 inches and the number of decks, n required to create uncertainty of 0.00002 inch is given by

$n=\frac {0.0001}{0.00002}=5$

We need 5 decks for the given uncertainty

Answer from: Quest

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