Guide n° 1 of solved exercises of units and notation
Solve the following exercises
problem #1
The time elapsed since the first animals inhabited the world, on dry land, is about 12,000,000,000,000,000 seconds. Express this time as a power of ten with a single figure, what is the order of magnitude?
problem #2
The speed of propagation of light in a vacuum is the same for all bodies and colors:
c = (2.99774 ± 0.00011)·10 5 km/s. what is the order of magnitude?
Problem #3
A ray of light takes approximately 1/100,000,000,000 seconds to pass through a window. How long does it take to go through a glass twice the size of the previous one? Compare the orders of magnitude of both times, how many glasses, like the first, must it go through, so that the order of magnitude changes?
Problem #4
Make the following conversions:
a) 24mg → kg
b) 8.6 cg → g
c) 2,600 dm³ → l
d) 92 cm³ → m³
e) 3 kg → g
f) 9 cm → m
g) 5h → s
h) 0.05 km → cm
i) 135 s → h
Problem #5
How many significant figures does each of the following numbers have?
a) 9
b) 90
c) 9,000.0
d) 0.009
e) 0.090
f) 909
g) 0.00881
h) 0.04900
i) 0.0224
i) 74.24
Problem #6
Express in a single number:
a) 3.59 10²
b) 4.32 10 -3
c) 3.05 10 -5
d) 5.29 10 5
e) 6.94 10¹
f) 0.05 10²
g) 1 10 8
h) 3.2 10 -3
i) 7.56 10 4
j) 0.00011 10 5
Problem #7
Carry out the following operations:
a) 1.29 10 5 + 7.56 10 4
b) 4.59 10 -5 – 6.02 10 -6
c) 5.4 10² 3.2 10 -3
Problem #8
Express in scientific notation:
a) 45.9
b) 0.0359
c) 45,967,800
d) 0.0005976
e) 345,690,000,000
f) 0.00011 10 5
Problem #9
How many significant figures should appear in the results of the following accounts?
a) 5 0.00559
b) 0.7 9.48 10¹
c) 875 67
| d) | 0.3 |
| 0.0586 |
| and) | 0.658 |
| 9.59 10¹ |
Solved problems of units and scientific notation
Statement of exercise n° 1
The time elapsed since the first animals inhabited the world, on dry land, is about 12,000,000,000,000,000 seconds. Express this time as a power of ten with a single figure, what is the order of magnitude?
Solution
12,000,000,000,000,000 s = 1.2 10 16 s
Order of magnitude: 17.
Statement of exercise n° 2
The speed of propagation of light in a vacuum is the same for all bodies and colors:
c = (2.99774 ± 0.00011) 10 5 km/s. what is the order of magnitude?
Solution
Order of magnitude: 6.
Statement of exercise n° 3
A ray of light takes approximately 1/100,000,000,000 seconds to pass through a window. How long does it take to go through a glass twice the size of the previous one? Compare the orders of magnitude of both times, how many glasses, like the first, must it go through, so that the order of magnitude changes?
Developing
Data:
t = 1/100,000,000,000 s = 0.00000000001 s = 1 10 -11 s
Order of magnitude: 12.
Solution
to)
Applying simple rule of three:
| 1 glass | → | 1 10 -11s |
| 2 glasses | → | t = 2 1 10 -11 s/1 |
t = 2 10 -11 s
Order of magnitude: 12, does not change.
b)
For the order of magnitude to vary, the “ten” must change, therefore it must go through at least 10 glasses like the first:
| 1 glass | → | 1 10 -11s |
| 10 glasses | → | t = 10 1 10 -11 s/1 |
t = 1 10 -10 s
Order of magnitude: 11.
Statement of exercise n° 4
Make the following conversions:
a) 24mg → kg
b) 8.6 cg → g
c) 2,600 dm³ → l
d) 92 cm³ → m³
e) 3 kg → g
f) 9 cm → m
g) 5h → s
h) 0.05 km → cm
i) 135 s → h
Solution
To convert units simply multiply and divide the “measure” (numerical value) by the unit expressed in both “magnitudes”.
a) We multiply by 1 kg and divide by the equivalent 1,000,000 mg:
24 mg (1 kg/1,000,000 mg) = 0.000024 kg = 2.4 10 -5 kg
b) We multiply by 1 g and divide by the equivalent 100 cg:
8.6 cg · (1 g/100 cg) = 0.086 g = 8.6 · 10 -2 g
c) We multiply by 1 l and divide by the equivalent 1 dm³:
2,600 dm³ (1 l/1 dm³) = 2,600 l = 2.6 10³ l
d) We multiply by 1 m³ and divide by the equivalent 1,000,000 cm³:
92 cm³ (1 m³/1,000,000 cm³) = 0.000092 m³ = 9.2 10 -5 m³
e) We multiply by 1,000 g and divide by the equivalent 1 kg:
3 kg·(1,000 g/1 kg) = 3,000 g = 3·10³ g
f) We multiply by 1 m and divide by the equivalent 100 cm:
9 cm (1 m/100 cm) = 0.09 m = 9 10 2 m
g) We know that 1 = 60 minutes and that one minute is equal to 60 seconds, therefore:
5 h = 5 h (60 m/1 h) = 300 m = 300 m (60 s/1 m) = 18,000 s = 1.8 10 4 s
h) We multiply by 100,000 cm and divide by the equivalent of 1 km:
0.05 km (100,000 cm/1 km) = 5,000 cm = 5 10³ cm
i) We know that 1 = 60 minutes and that one minute is equal to 60 seconds, therefore:
135 s = 135 s (1 m/60 s) = 2.25 m = 2.25 m (1 h/ 60m) = 0.0375h = 3.75 10 -2h
Statement of exercise n° 5
How many significant figures does each of the following numbers have?
a) 9
b) 90
c) 9,000.0
d) 0.009
e) 0.090
f) 909
g) 0.00881
h) 0.04900
i) 0.0224
i) 74.24
Solution
| figures | |
| a) 9
b) 90 c) 9,000.0 d) 0.009 e) 0.090 f) 909 g) 0.00881 h) 0.04900 i) 0.0224 i) 74.24 |
one
one 5 one two 3 3 4 3 4 |
Statement of exercise n° 6
Express in a single number:
a) 3.59 10²
b) 4.32 10 -3
c) 3.05 10 -5
d) 5.29 10 5
e) 6.94 10¹
f) 0.05 10²
g) 1 10 8
h) 3.2 10 -3
i) 7.56 10 4
j) 0.00011 10 5
Solution
a) 3.59 10² = 3.59 100 = 359
b) 4.32 10 -3 = 4.32/1,000 = 0.00432
c) 3.05 10 -5 = 3.05/100,000 = 0.0000305
d) 5.29 10 5 = 5.29 100,000 = 529,000
e) 6.94 10¹ = 6.94 10 = 69.4
f) 0.05 10² = 0.05 100 = 5
g) 1 10 8 = 1 100,000,000 = 100,000,000
h) 3.2 10 -3 = 3.2/1,000 = 0.0032
i) 7.56 10 4 = 7.56 10,000 = 75,600
j) 0.00011 10 5 = 0.00011 100,000 = 11
Statement of exercise No. 7
Carry out the following operations:
a) 1.29 10 5 + 7.56 10 4
b) 4.59 10 -5 – 6.02 10 -6
c) 5.4 10² 3.2 10 -3
Solution
a) 1.29 10 5 + 7.56 10 4 = 129,000 + 75,600 = 204,600 = 2.046 10 5
b) 4.59 10 -5 – 6.02 10 -6 = 0.0000459 – 0.00000602 = 0.00003988 = 3.988 10 -5
c) 5.4 10² 3.2 10 -3 = (5.4 3.2) 10 (2-3) = 17.28 10 -1 = 1.728
Statement of exercise n° 8
Express in scientific notation:
a) 45.9
b) 0.0359
c) 45,967,800
d) 0.0005976
e) 345,690,000,000
f) 0.00011 10 5
Solution
a) 45.9 = 4.59 10
b) 0.0359 = 3.59 10 -2
c) 45,967,800 = 4.59678 10 7
d) 0.0005976 = 5.976 10 -4
e) 345,690,000,000 = 3.4569 10 11
f) 0.00011 10 5 = 1.1 10 -4 10 5 = 1.1 10
Statement of exercise n° 9
How many significant figures should appear in the results of the following accounts?
a) 5 0.00559
b) 0.7 9.48 10¹
c) 875 67
d) 0.3/0.0586
e) 0.658/9.59 10¹
Solution
a) 5 0.00559 = 0.02795 → 4 significant figures
b) 0.7 9.48 10¹ = 0.7 9.48 10 = 66.36 → 4 significant figures
c) 875 67 = 58,625 → 5 significant figures
d) 0.3/0.0586 = 5.1194539249146757679180887372014 ≈ 5.1195 → 5 significant figures
e) 0.658/9.59 10¹ = 0.658/9.59 10 = 0.00686131386861313868613138686131 ≈ 0.00686 → 3 significant figures
