Chapter 4.7, Problem 52E

a.

To determine

To find scientific notations of $0.000003$ meter, $0.0000022$ meter and the combined length of a pit and a land.

Answer to Problem 52E

  $3\times {10}^{-6}meter,2.2\times {10}^{-6}meter,5.2\times {10}^{-6}meter$

Explanation of Solution

Given information: $0.000003$ meter, $0.0000022$ are the lengths of land and pit respectively.

Formula used:using the scientific notations of the said figures.

Calculation: $0.000003$ meter can be written as $3*{10}^{-6}meter$

  $0.0000022$ meter can be written as $2.2*{10}^{-6}meter$

Combined length of a pit and a land is:

  $0.000003+0.0000022$

  $=0.0000052$

  $=5.2\times {10}^{-6}meter$

b.

To determine

To find the length of series of pits and lands be if the disc containing $2,000,000,000$ pits and lands.

Answer to Problem 52E

  $1.1\times {10}^4$ meter

Explanation of Solution

Given information: $0.000003$ meter, $0.0000022$ are the lengths of land and pit respectively. Disc contains $2,000,000,000$ each of pits and lands

Formula used: using the scientific notations of the said figures. For multiplication, used the product of powers.

Calculation: $0.000003$ meter can be written as $3\times {10}^{-6}meter$

  $0.0000022$ meter can be written as $2.2\times {10}^{-6}meter$

  $2,000,000,000$ can be written as $2\times {10}^9$

Combined length of a pit and a land is:

  $0.000003+0.0000022$

Adding the two terms

  $=0.0000052$

  $=5.2\times {10}^{-6}meter$

Total length of $2\times {10}^9$ such pits and lands:

  $2\times {10}^9\left(5.5\times {10}^{-6}\right)$ meter

  $=2\times 5.5\times {10}^{9-6}$

Using the product of powers rule

  $=11\times {10}^3$

Changing to scientific notation.a

  $=1.1\times {10}^4$ meter in scientific notation.