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Chapter 1: Problem 16

Express each value in exponential form. Where appropriate, include units in your answer. (a) solar radiation received by Earth: 173 thousand trillion watts (b) average human cell diameter: 1 ten-millionth of a meter (c) the distance between the centers of the atoms in silver metal: 142 trillionths of a meter (d) $\frac{\left(5.07 \times 10^{4}\right) \times\left(1.8 \times 10^{-3}\right)^{2}}{0.065+\left(3.3 \times 10^{-2}\right)}=$

Short Answer

Expert verified
\n(a) Solar radiation: \(1.73 \times 10^{26}\) watts \n(b) Average human cell diameter: \(1.0 \times 10^{-7}\) meters \n(c) Distance between atoms in Silver: \(1.42 \times 10^{-10}\) meters \n(d) Computed Value = \(7.28 \times 10^{10}\)

Step by step solution

01

Express Solar Radiation

Given Solar radiation received by Earth: 173 thousand trillion watts. We will express this value in exponential format giving the answer as \(1.73 \times 10^{26}\) watts
02

Express Average Human Cell Diameter

Given Average human cell diameter: 1 ten-millionth of a meter. We express this value in exponential format giving the result as \(1.0 \times 10^{-7}\) meters
03

Express Distance between Atoms in Silver

Given Distance between the centers of atoms in silver metal: 142 trillionths of a meter. Converting this into exponential notation, we get \(1.42 \times 10^{-10}\) meters
04

Evaluate the Given Expression

We will simplify the given expression \(\frac{\left(5.07 \times 10^{4}\right) \times\left(1.8 \times 10^{-3}\right)^{2}}{0.065+\left(3.3 \times 10^{-2}\right)}\), by performing operations in the order recommended by the order of operations, BEDMAS (Brackets, Exponents, Division, Multiplication, Addition, Subtraction). After performing the operations in the correct order, we get \(7.28 \times 10^{10}\)

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Most popular questions from this chapter

The fact that the volume of a fixed amount of gas at a fixed temperature is inversely proportional to the gas pressure is an example of (a) a hypothesis; (b) a theory; (c) a paradigm; (d) the absolute truth; (e) a natural law.

Calculate the mass of a block of iron \(\left(d=7.86 \mathrm{g} / \mathrm{cm}^{3}\right)\) with dimensions of \(52.8 \mathrm{cm} \times 6.74 \mathrm{cm} \times 3.73 \mathrm{cm}\).

In the United States, volume of irrigation water is usually expressed in acre- feet. One acre-foot is a volume of water sufficient to cover 1 acre of land to a depth of 1 ft \(\left(640 \text { acres }=1 \mathrm{mi}^{2} ; 1 \mathrm{mi}=5280 \mathrm{ft}\right)\) The principal lake in the California Water Project is Lake Oroville, whose water storage capacity is listed as \(3.54 \times 10^{6}\) acre-feet. Express the volume of Lake Oroville in (a) cubic feet; (b) cubic meters; (c) U.S. gallons.

Express each of the following to fur significant figures. (a) 3984.6 (b) 422.04 (c) 0.0033 (d) 902.10 (e) 0.02173 (f) 7000 (g) 7.02 (h) 67,000,000

To determine the approximate mass of a small spherical shot of copper, the following experiment is performed. When 125 pieces of the shot are counted out and added to \(8.4 \mathrm{mL}\) of water in a graduated cylinder, the total volume becomes \(8.9 \mathrm{mL}\). The density of copper is \(8.92 \mathrm{g} / \mathrm{cm}^{3} .\) Determine the approximate mass of a single piece of shot, assuming that all of the pieces are of the same dimensions.
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