14. How many a. 6.21 b. 62.1 c. 6210 d. 6210.0 3 3 e. 0.0621 f,0.620 g. 0.623 h.

Question

14. How many a. 6.21 b. 62.1 c. 6210 d. 6210.0 3 3 e. 0.0621 f,0.620 g. 0.623 h. .62 j. 6.21 × 103 k. 6.21 × 10-3 3 1.621 × 103 3 ing the significant figure standards adopted for this text. 15. Compute the following numbers, apply a 33.3 x 254-815 b. 33.3 c, 33.3 ÷ 45.1)= d, 33.3 × 45-1, e. 2.345 × 3.321 f.(4.32 × 1.23)-(5)- .33.32 aly 8.81 25.4= . .738 16. Express the following numbers and computed results in scientific notation, paying attention to significant figures 9.3rM d. 32,014 × 47 = 1,510 S: ylos e. 0.059 ÷ 2.304 = 2.56 ×10-5 f. 320. x 0050- 1.Gx1O a. 9,827= b. 0000000550= 5.5x10- e. 3,200.000 3a XIou 17. Convert the following to SI units, Work across the line and show all teps in the conversion Use scientific notation and apply the proper use of significant figures. Note: Think carefully about g and h. Pictures may help. a. 9.12us × b. 3.42 km 44 cm/ms d. 80 km/hr× e. 60 mphx f. 8in× c, × 14 in3x h.250 cm1x

Explanation / Answer

Given

A. 9.12 micro seconds

we know that in SI system the fundamental quantities are  

for mass is in kg,length is in meter , time is in seconds

so 9.12 micro seconds , micro means 10^-6 part of a quantity

9.12*10^-6 seconds

b. 3.42 km

we know one km = 1000 m

3.42*1000 m = 3.42*10^3 m

c. 44 cm /ms

converting cm in to meters is 1 cm = 10^-2 m

and converting ms in to seconds is 1 ms = 10^-3 s

so 44 cm/ms = 44((10^-2)/10^-3) = 440 m /s

d. 80 km/hr

converting km in to meters is 1 km = 10^3 m

and converting hr in to seconds is 1 hr = 3600 s

so 80 km/hr = 80*1000/3600 m/s = 22.22 m/s

e. 60 mph

we know one mile = 1600 m

and one hour = 3600 s

so 60 mph = 60*1600/3600 m/s = 26.667 m/s

f.

8 in

we know one inch = 0.0254 m

so 8 in = 8*0.0254 m = 0.2032 m

g. 14 in^2

we know that 1 in^2 = 0.00064516 m^2

14 in^2 = 14*0.00064516 m^2 = 0.00903224 m^2

h. 250 cm^3

we know that 1 cm^3 = 10^-6 m^3

so 250 cm^3 = 250*10^-6 m^3 = 0.00025 m^3

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