완도군청

대학교재솔루션 다운로드 물리학 다운로드 할리데이 일반물리학 8판 대학교재솔루션 haliday 일반물리학 Halliday Physics 8ed solution 다운받기


자료보러가기 ▶▶▶▶▶ http://my.allreport.co.kr/nfalle/11023483.html


대학교재솔루션 다운로드 물리학 다운로드 할리데이 일반물리학 8판 대학교재솔루션 haliday 일반물리학 Halliday Physics 8ed solution

할리데이 일반물리학 8판 솔루션 총 분량은 4130 페이지 영문판입니다.

할리데이 일반물리학 8판 솔루션
총 분량은 4130 페이지

영문판입니다.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 μm,

1km = 103 m = 103 m 106 μ m m = 109 μ m. The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 × 109 μm. (b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10?2 m, 1cm = 10?2 m = 10?2 m 106 μ m m = 104 μ m. We conclude that the fraction of one centimeter equal to 1.0 μm is 1.0 × 10?4. (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
1.0 yd = ( 0.91m ) 106 μ m m = 9.1 × 105 μ m.

(

)(

)

(

)(

)

(

)

2. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain 1 inch 6 picas ? 1.9 picas. 0.80 cm = ( 0.80 cm ) 2.54 cm 1 inch (b) With 12 points = 1 pica, we have
0.80 cm = ( 0.80 cm ) 1 inch 2.54 cm 6 picas 1 inch 12 points ? 23 points. 1 pica

3. Using the given conversion factors, we find (a) the distance d in rods to be
d = 4.0 furlongs =

( 4.0 furlongs )( 201.168 m furlong )
5.0292 m rod

= 160 rods,

(b) and that distance in chains to be
d =

( 4.0 furlongs )( 201.168 m furlong )
20.117 m chain

= 40 chains.

4. The conversion factors 1 gry = 1/10 line , 1 line=1/12 inch and 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 0.50 gry 2 = 0.18 point 2 .

5. Various geometric formulas are given in Appendix E. (a) Expressing the radius of the Earth as

R = ( 6.37 × 106 m )(10?3 km m ) = 6.37 × 103 km,
its circumference is s = 2π R = 2π (6.37 × 103 km) = 4.00 ×104 km. (b) The surface area of Earth is A = 4π R 2 = 4π ( 6.37 × 103 km ) = 5.10 × 108 km 2 .
2

(c) The volume of Earth is V =

4 π 3 4π R = 6.37 × 103 km 3 3

(

)

3

= 1.08 × 1012 km3 .

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
50.0 S = ( 50.0 S) 258 W = 60.8 W 212 S

(b) In units of Z, we have
50.0 S = ( 50.0 S ) 156 Z = 43.3 Z 180 S

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 π V = r2 z 2 where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have
r = ( 2000 km )
103 m 1km 102 cm = 2000 × 105 cm. 1m

In these units, the thickness becomes
z = 3000 m = ( 3000 m )

102 cm = 3000 × 102 cm 1m
2

which yields V =

π 2000 × 105 cm 2

(

) ( 3000 × 10
2

cm = 1.9 × 1022 cm3 .

)

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. 1 Thus, 1 fanega = 12 cah


자료출처 : http://my.allreport.co.kr/nfalle/11023483.html



[문서정보]

문서분량 : 4,130 Page
파일종류 : PDF 파일
자료제목 :
파일이름 : 솔루션 물리 일반물리학8판halliday.pdf
키워드 : PEET,MEET,DEET,할리데이,물리학,솔루션,8판,Halliday,Physics,8ed



[관련자료]

- 솔루션 일반물리학8판 북스힐
- [솔루션] 할리데이 일반물리학 9판 솔루션 (Halliday - Fundamentals of Physics 9e solutions)
- [솔루션]일반물리학 8판 할리데이 haliday 8ed 솔루션
- [솔루션] 일반 물리학 9판 솔루션(Fundamentals of Phyics 9E Solutions Manual, Halliday - Wiley)
- [대학교재 솔루션] 응용선형대수 솔루션 (Gilbert strang Linear Algebra and its application 4e) [gilbert strang] linear algebra and its application
- [대학교재 솔루션] 유기화학 7판 Susan McMurry(맥머리) 솔루션 맥머리(Susan McMurry)유기화학7판
- [물리솔루션] 할리데이 일반물리학 8판 솔루션 haliday 8ed
- [솔루션] 대학물리학7판 - PHYSICS 7th JEWETT SERWAY THOMSON Brooks&Core (ch1-22장)
- [솔루션] 미분적분학 교우사 솔루션


자료보러가기 ▶▶▶▶▶ http://my.allreport.co.kr/nfalle/11023483.html