ANSWER TO CHAPTER 1 HOMEWORK

1. (a) Assuming one significant figure, we have

                10 billion yr = 10 x 109 yr =  1 x 1010 yr

        (b)  (1 x1010 yr)(3 x 107 s/yr) = 3 x 1017 s.

2. (a) 4 significant figures.

        (b) Because athe zero is not needed for placement, we have 4 significant figures.   

        (c) 3 significant figures.

        (d) Because the zeros are for placement only, we have 1 significant figure.

        (e) Because the zeros are for placement only, we have 2 significant figures.

        (f) 4 significant figures.

        (g) 2, 3, or 4 significant figures,   depending on the significance of the zeros.

3. (a) 1,156 = 1.156 x 103.

        (b) 21.8 = 2.18 x 101

        (c) 0.0068 = 6.8 x 10-3.

        (d) 27.635 = 2.7635 x101.

        (e) 0.219 = 2.19 x 10-1.

        (f) 22 = 2.2 x 101.

4. (a) 8.69 x 104 = 86,900.

        (b) 7.1 x 103 = 7,100.

        (c) 6.6 x 10-1 = 0.66

        (d) 8.76 x 102 = 876.

        (e) 8.62 x 10-5 = 0.000 0862.

5. % uncertainty = [(0.25m)/(3.26m)] 100 = 7.7%

        Because the uncertainty  has 2 significant figures,  the % uncertainty has 2 significant figures.

6. For multiplication, the number of significant figures in the result is the least number from the multipliers; in this case 2 from the second value.

             (2.079 x 102m)(0.072 x 10-1) = 0.15 x 101m = 1.5m.

7. To add, we make all of the exponents the same:

             9.2 x 103s + 8.3 x 104s + 0.008 x 106s = 0.92 x 104s + 8.4 x 104s + 0.8 x 104s

                  = 10.02 x 104s = 1.0 x 105s.

         Because we are adding, the location of the uncertain figure for the result is the one furthest to the left. In this case, it is fixed by the third value.

8. (a) 106 volts = 1 megavolt = 1 Mvolt.

        (b) 10-6 meters = 1 micrometer = 1 mm.

        (c) 6 x 103 days = 6 kilodays = 6 kdays.

        (d) 18 x 102 bucks = 18 hectobucks = 1.8 kbucks.

        (e) 8 x 10-9 pieces = 8 nanopieces = 8 npieces.

9. (a) 286.6 mm = 286.6 x 10-3 m = 0.286 6 m

        (b) 85 mV = 85 x 10-6 V = 0.000 085 V.

        (c) 760 mg = 760 x 10-3g = 760 x 10-6 kg = 0.000 760 kg. This assumes that the last zero is significant.

          (d) 60.0 picoseconds = 60.0 x 10-12s = 0.000 000 000 060 0 s.

        (e) 22.5 femtometers = 22.5 x 10-15 m = 0.000 000 000 000 022 5 m.

        (f) 2.50 gigavolts = 2.50 x 109 volts = 2,500,000,000 volts.

10  (a) 2800 = 2.8 x 103  »  1x 103 = 103.

          (b) 86.30 x 102 » 100 x 102 = 104.

          (c) 0.0076 = 0.76 x 10-2 » 10-2.

          (d) 15.0 x 108 = 1.50 x 109 » 109.

 
HOME SEND QUESTIONS