1,079,252,848 km/h or 299,792,457.8m/s
10 figure of speed
0.5 light speed
4 significant figures. once put in scientific notation you will see 2.057 X 10^-2 You will ignore the last zero since it is to the farthest right past the decimal and not necessary.
a 10 speed racer is a dirt bike a 10 speed racer is a dirt bike
First write out the problem: 3.27*108= x * 6.43 * 1017 Divide both sides by 6.43 * 1017 (3.27/6.43) * 108-17=x .509 * 10-11=x (Notice that I rounded it to 3 significant figures) If you want to put it in scientific notation then the answer is 5.09 * 10-12
The Radius of the Star is 10 KM. Now we need to find the diameter. Circumference equals pi times the Diameter. That is twice the Radius, which is 20. Now we need to get the Diameter. We multiply the diameter by pi. 20 times 3.1416. 20 * 3.1416 = 62.838. Now we multiply the circumference in kilometers by the speed. 62.832 * 642 = 40,338. Now we take the speed of light, 299,997 per second and divide it into the speed of a point on the equator of our star going 40,338 kilometers per second. 40338 / 299997 = 0.1345
10 significant figures.
10 significant figures.
10 significant figures.
There are two significant figures which are the two 2s.
The speed of light through a vacuum is exactly 299,792,458 m/s. However, it is usually given as3.00 x 10^8 m/s to three significant figures.
Ten of them.
3 significant figures. 8.53ร10ยน.
10 significant figures.
Significant figures are very important when it comes to calculations. If the mass of an electron is 9.10939 x 10-31 then its significant figures are: 9 x 10^-31( correct 1 significant figure), 9.1 x 10^-31 kg ( correct to 2 significant figures), 9.11 x 10^-31 (correct to 3 significant figures), and 9.109 x 10^-31 (correct to 4 significant figures).
600 * 190 = 114*10+3 in three significant figures, or 1.14 * 105 also in three significant figures.
How many significant figures are in 0.074100x 10^-4
10 significant figures.