The distance of the earth from the sun is 149,000,000 km. In scientific notation the distance is:

149 \times 10^{6}

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14.9 \times 10^{7}

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1.49 \times 10^{8}

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0.149 \times 10^{9}

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Solution

The correct option is C $1.49 \times 10^{8}$ km
$149, 000, 000 k m = 149 \times 1000000 = 1.49 \times 100000000 = 1.49 \times 10^{8} k m$

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Similar questions

A light year is the distance that light can travel in one year.
$1$ light year $= 9, 460, 000, 000, 000 k m .$
(a) Express one light year in scientific notation.
(b) The average distance between Earth and Sun is $1.496 \times 10^{8} k m .$ Is the distance between earth and Sun greater than, less than or equal to one light year?

Q. The mean distance of Earth from the Sun is

149.6 \times 10^{6}

km and the mean distance of Mercury from the Sun is

57.9 \times 10^{6}

km. The period of Earth’s revolutions is 1 year, what is the period of Mercury’s revolution?

Q. The average distance of the sun from the earth is about

a. 15 crore km.
b. 1.5 crore km.
c. 150 crore km.
d. 1,500 crore km.

Q. Earth revolves around the sun in 365 days. If the average distance from the sun is 150 million km. Find earth's linear velocity.

Q. Investigating Solar System The table shows the average distance from each planet in our solar system to the sun.

Celestial Body	Distance $(k m)$	Distance $(k m)$ Standard Notation
Earth	$149, 600, 000$	$1.496 \times 10^{8}$
Jupiter	$778, 300, 000$
Mars	$227, 900, 000$
Mercury	$57, 900, 000$
Neptune	$4, 497, 000, 000$
Pluto	$5, 900, 000, 000$
Saturn	$1, 427, 000, 000$
Uranus	$2, 870, 000, 000$
Venus	$108, 200, 000$

Complete the table by expressing the distance from each planet to the Sun in scientific notation.

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The distance of the earth from the sun is 149,000,000 km. In scientific notation the distance is:

The correct option is C 1.49×108 km149,000,000km=149×1000000=1.49×100000000=1.49×108km

The correct option is C $1.49 \times 10^{8}$ km
$149, 000, 000 k m = 149 \times 1000000 = 1.49 \times 100000000 = 1.49 \times 10^{8} k m$