Modified Mastering Astronomy with Pearson eText -- Standalone Access Card -- for Astronomy Today (9th Edition)
Modified Mastering Astronomy with Pearson eText -- Standalone Access Card -- for Astronomy Today (9th Edition)
9th Edition
ISBN: 9780134553955
Author: Eric Chaisson, Steve McMillan
Publisher: PEARSON
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Chapter 1, Problem 2P

(a)

To determine

The scientific notation forms of the numbers 1000; 0.000001;  1001; 1,000,000,000,000,000; 123,000; 0.000456.

(a)

Expert Solution
Check Mark

Explanation of Solution

Formula Used:

A number in scientific notation is in form a×10n, where 1|a|<10 and n is integer.

Explanation:

Consider the number 1000. It is greater than 10. So, positive power of 10 is used. Then, the number 1000 in scientific notation, that is in the form a×10n, will be 1×103.

Now, consider the number 0.000001. The number is in between 0 and 1. So, negative power of 10 is used in the scientific notation. Then, the number 0.000001 in scientific notation will be 1×106.

Similarly, the number 1001 will be written in scientific notation as 1.001×103, the number 1,000,000,000,000,000 as 1.00×1015, and the number 123,000 as 1.23×105, since the number is greater than 10.

But, the number 0.000456 will be written in scientific notation as 4.56×104 since the number is between 0 and 1.

(b)

To determine

To write: The numbers 3.16×107; 2.998×105;  6.67×1011; and 2×100 in the “normal” numerical form.

(b)

Expert Solution
Check Mark

Explanation of Solution

Formula Used:

A normal numerical form is the one in which the number is not written in the exponent form.

Explanation:

Consider the number 3.16×107. In the “normal” numerical form, it will be written as,

3.16×107=3.16×10,000,000=316100×10,000,000=31,600,000

Similarly, consider the number 2.998×105. In the “normal” numerical form, it will be written as,

2.998×105=2.998×100,000=29981000×100,000=299,800

And, the number 6.67×1011 in the “normal” numerical form will be,

6.67×1011=6.67100,000,000,000=0.0000000000667

And, the number 2×100 in the “normal” numerical form will be,

2×100=2×1=2

(c)

To determine

The value of the expressions (2×103)+102;(1.99×1030)/(5.98×1024), and (3.16×107)×(2.998×105).

(c)

Expert Solution
Check Mark

Explanation of Solution

Formula Used:

Rule for the multiplication of the numbers with exponents: Multiply the factors and add the exponents.

Rule for the division of the numbers with exponents: Divide the factors and subtract the exponents.

Explanation:

Consider the expression (2×103)+102.

For adding the two numbers, both should have same exponents. Here, the exponents are not same thus converting both the numbers in normal numerical form and then adding,

(2×103)+102=2000+0.01=2000.01

Now, consider the expression (1.99×1030)/(5.98×1024).

For dividing the two numbers, divide the factors and subtract the exponents.

(1.99×1030)/(5.98×1024)=(1.995.98)×103024=0.332×106

Consider the expression (3.16×107)×(2.998×105).

For multiplying the two numbers, multiply the factors and add the exponents.

(3.16×107)×(2.998×105)=(3.16×2.998)×107+5=9.47×1062

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