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Chapter 1: Q15. (page 18)

The sun, on average, is 93 million miles from the earth. How many meters is this? Express (a) using powers of 10, and (b) using a metric prefix (km).

Short Answer

Expert verified

The results obtained for parts (a) and (b) are $d = 1.49 \times 10^{11} m$ and $d = 1.49 \times 10^{8} km$ .

Step by step solution

01

Step 1. Converting miles into meters

Mile is a unit of measurement of distance. One mile is equivalent to 5280 feet.

Given data:

Distance of the sun from the earth, $d = 93 million - miles$

(a)

Using the formula to express a million miles in meters, you get

$\begin{matrix} d = (93 million - miles \times \frac{10^{6} miles}{1 million - miles}) \\ = (93 \times 10^{6} miles \times \frac{1609.34 meters}{1 mile}) \\ = 1.49 \times 10^{11} m . \end{matrix}$

02

Step 2. Converting meters into kilometers

For converting meters into kilometers, divide the given distance by a factor of 1000.

(b)

Using the formula to determine the distance with the metric prefix, you get

$\begin{matrix} d = (1.49 \times 10^{11} m \times \frac{1 km}{1000 m}) \\ = 1.49 \times 10^{8} km . \end{matrix}$

Thus, the values of distance in powers of 10 and in the metric prefix are $1.49 \times 10^{11} m$ and $1.49 \times 10^{8} km$ , respectively.

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Most popular questions from this chapter

Which is not true about an order-of-magnitude estimation?

(a) It gives you a rough idea of the answer.

(b) It can be done by keeping only one significant figure.

(c) It can be used to check if an exact calculation is reasonable.

(d) It may require making some reasonable assumptions in order to calculate the answer.

(e) It will always be accurate to at least two significant figures.

Make a rough estimate of the volume of your body (in $m^{3}$ ).

How big is a ton? That is, what is the volume of something that weighs a ton? To be specific, estimate the diameter of a 1-ton rock, but first make a wild guess: will it be 1 ft across, 3 ft, or the size of a car? [Hint: Rock has mass per volume about 3 times that of water, which is 1 kg per liter $(10^{3} {cm}^{3})$ or 62 lb per cubic foot.]

A watch manufacturer claims that its watches gain or lose no more than 8 seconds in a year. How accurate are these watches, expressed as a percentage?

Determine the percent uncertainty in $θ$ ,and in $\sin θ$ , when (a) $θ = 15 ° \pm 0.5 °$ , (b) $θ = 75 ° \pm 0.5 °$ .

See all solutions

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