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Chapter 1: Problem 72

A copper wire (density \(=8.96 \mathrm{~g} / \mathrm{cm}^{3}\) ) has a diameter of \(0.25 \mathrm{~mm}\). If a sample of this copper wire has a mass of \(22 \mathrm{~g}\), how long is the wire?

Short Answer

Expert verified
The length of the copper wire is approximately 4998.88 cm or about 49.99 meters.

Step by step solution

01

Find the volume of the wire

We are given the mass as 22 g and the density as 8.96 g/cm鲁. We can use the formula for density to find the volume: \[density = \frac{mass}{volume}\] From this formula, we can find the volume: \[volume = \frac{mass}{density}\] Substitute the given values into the formula: \[volume = \frac{22}{8.96}\] Now, calculate the volume: \[volume \approx 2.4536 \, \mathrm{cm}^{3}\] So, the volume of the wire is approximately 2.4536 cm鲁.
02

Find the cross-sectional area of the wire

We are given the diameter (0.25 mm) of the wire. First, convert the diameter from mm to cm: \[diameter = 0.25 \, \mathrm{mm} \times \frac{1 \, \mathrm{cm}}{10\, \mathrm{mm}} = 0.025 \, \mathrm{cm}\] Now find the radius: \[radius = \frac{diameter}{2} = \frac{0.025}{2} = 0.0125 \, \mathrm{cm}\] The cross-sectional area of the wire is the area of a circle, which can be found using the formula: \[A = \pi r^2\] Substitute the radius value into the formula: \[A = \pi\times (0.0125)^2\] Now, calculate the cross-sectional area: \[A \approx 0.00049087 \, \mathrm{cm}^{2}\] So, the cross-sectional area of the wire is approximately 0.00049087 cm虏.
03

Find the length of the wire

Now, we can use the volume and the cross-sectional area to find the length of the wire. Divide the volume of the sample by the cross-sectional area of the wire: \[length = \frac{volume}{cross-sectional\, area}\] Substitute the given values into the formula: \[length = \frac{2.4536}{0.00049087}\] Now, calculate the length: \[length \approx 4998.8824 \, \mathrm{cm}\] So, the length of the wire is approximately 4998.88 cm or about 49.99 meters.

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Most popular questions from this chapter

The world record for the hundred meter dash is \(9.74 \mathrm{~s}\). What is the corresponding average speed in units of \(\mathrm{m} / \mathrm{s}, \mathrm{km} / \mathrm{h}, \mathrm{ft} / \mathrm{s}\), and \(\mathrm{mi} / \mathrm{h}\) ? At this speed, how long would it take to run \(1.00 \times 10^{2}\) yards?

If a piece of hard white blackboard chalk is heated strongly in a flame, the mass of the piece of chalk will decrease, and eventually the chalk will crumble into a fine white dust. Does this change suggest that the chalk is composed of an element or a compound?

Evaluate each of the following and write the answer to the appropriate number of significant figures. a. \(212.2+26.7+402.09\) b. \(1.0028+0.221+0.10337\) c. \(52.331+26.01-0.9981\) d. \(2.01 \times 10^{2}+3.014 \times 10^{3}\) e. \(7.255-6.8350\)

According to the Official Rules of Baseball, a baseball must have a circumference not more than \(9.25\) in or less than \(9.00\) in and a mass not more than \(5.250 \mathrm{z}\) or less than \(5.00 \mathrm{oz}\). What range of densities can a baseball be expected to have? Express this range as a single number with an accompanying uncertainty limit.

When a marble is dropped into a beaker of water, it sinks to the bottom. Which of the following is the best explanation? a. The surface area of the marble is not large enough to be held up by the surface tension of the water. b. The mass of the marble is greater than that of the water. c. The marble weighs more than an equivalent volume of the water. d. The force from dropping the marble breaks the surface tension of the water. e. The marble has greater mass and volume than the water. Justify your choice, and for choices you did not pick, explain what is wrong about them.
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