Desenho Tecnico

1. We use Eq. 2-2 and Eq. 2-3. During a time tc when the velocity remains a positive constant, speed is equivalent to velocity, and distance is equivalent to displacement, with Δx = v tc. (a) During the first part of the motion, the displacement is Δx1 = 40 km and the time interval is t1 40 133 == ( . km) (30 km/ h) h.

During the second part the displacement is Δx2 = 40 km and the time interval is

t2

40

067 == ( . km) (60 km/ h) h.

Both displacements are in the same direction, so the total displacement is

Δx = Δx1 + Δx2 = 40 km + 40 km = 80 km.

The total time for the trip is t = t1 + t2 = 2.00 h. Consequently, the average velocity is

vavg

km) (2.0 h)1. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2).

(a) Since 1 km = 1 × 103 m and 1 m = 1 × 106 μ

m, ()() 3 3 6 9 1km 10 m 10 m 10 m m 10 m. == = μμ

The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 × 109 μ m.

(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10−2 m, ()() 2 2 6 4 1cm = 10 m = 10 m 10 m m 10 m. −− = μμ

We conclude that the fraction of one centimeter equal to 1.0 μ

m is 1.0 × 10−4.

(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m, ()() 65 1.0yd = 0.91m 10 m m 9.1 10 m. =× μμ

2. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain ()1inch 6 picas0.80 cm = 0.80 cm 1.9 picas. 2.54 cm 1inch § ·§ · ≈ ¨ ¸¨ ¸ © ¹© ¹

(b) With 12 points = 1 pica, we have

()1inch 6 picas 12 points0.80 cm = 0.80 cm 23 points. 2.54 cm 1inch 1 pica § ·§ ·§ · ≈ ¨ ¸¨ ¸¨ ¸ © ¹© ¹© ¹

(b) and that distance in chains to be ()() 4.0 furlongs 201.168 m furlong 40 chains. 20.117 m chain d ==

3. Using the given conversion factors, we find

(a) the distance d in rods to be

()() 4.0 furlongs 201.168 m furlong 4.0 furlongs = 160 rods, 5.0292 m rod d ==

4. The conversion factors 1 gry 1/10 line = , 1 line=1/12 inchand 1 point = 1/72 inch imply that

1 gry = (1/10)(1/12)(72 points) = 0.60 point.

Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 22 0.50 gry = 0.18 point .

5. Various geometric formulas are given in Appendix E.

(a) Expressing the radius of the Earth as () () 63 3 6.37 10 m 10 km m 6.37 10 km,R −=× = ×

its circumference is 34 2 2 (6.37 10 km) 4.00 10 km. sR ππ = = × = × (b) The surface area of Earth is () 2 2 3 8 2 4 4 6.37 10 km 5.10 10 km . AR = π = π × = × (c) The volume of Earth is () 3 3 3 12 3 44 6.37 10 km 1.08 10 km . 33 VR ππ = = × = ×

6. From Figure 1.6, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z.

(a) In units of W, we have

() 258 W 50.0 S 50.0 S 60.8 W 212 S §· == ¨¸ ©¹

(b) In units of Z, we have

() 156 Z 50.0 S 50.0 S 43.3 Z 180 S §· == ¨¸ ©¹

2

2 V r z π=

where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have

() 32 510 m 10 cm2000km 2000 10 cm. 1km 1m

r

§ · § · ==× ¨ ¸ ¨ ¸ © ¹ © ¹

In these units, the thickness becomes () 2 210 cm3000m 3000m 3000 10 cm 1m z §· == = × ¨¸ ©¹ which yields ()() 2 5 2 22 3 2000 10 cm 3000 10 cm 1.9 10 cm . 2 V π = × × = ×

7. The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = πr2/2, where r is the radius. Therefore, the volume is

8. We make use of Table 1-6.

(a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen fanega. Thus, 1 fanega = 1 12 cahiz, or 8.33 × 10−2cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = 1 48 cahiz, or 2.08 × 10−2 cahiz. Continuing in this way, the remaining entries in the first column are 6.94 × 10−3 and 33.47 10 −× . (b) In the second (“fanega”) column, we similarly find 0.250, 8.33 × 10−2, and 4.17 × 10−2 for the last three entries.

(c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries.

(d) Finally, in the fourth (“almude”) column, we get 1 2 = 0.500 for the last entry.

(e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.00 almudes must be equal to 14.0 medios.

(f) Using the value (1 almude = 6.94 × 10−3 cahiz) found in part (a), we conclude that 7.00 almudes is equivalent to 4.86 × 10−2 cahiz.

(g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 m3 or 55501 cm3. Thus, 7.00 almudes = 7.00 12 fanega = 7.00 12 (55501 cm3) = 3.24 × 104 cm3.

231 acre ft = (43,560 ft ) ft = 43,560 ft ⋅⋅

Since 2 in. = (1/6) ft, the volume of water that fell during the storm is

2 2 2 7 3 (26 km )(1/6 ft) (26 km )(3281ft/km) (1/6 ft) 4.66 10 ft .V == =×

Thus,

V = × ×⋅ = × ⋅466 10 43560 10 11 10 7 4 3. . . ft ft acre ft acre ft. 3 3

9. We use the conversion factors found in Appendix D.

10. A day is equivalent