Su'aalaha tusaalaha ah ee Vector

25 Tusaalooyin oo ah Dhibaatooyinka Vektorka

1. Laba ciidan ayaa isku toosan, mid walbana wuxuu leeyahay cabbir dhan 3 N iyo 4 N. Cabbirka ka dhashay labada ciidan waa...
Dood
Waa la ogyahay in:
F₁ = 3 N, F₂ = 4N
Su'aal: Waa maxay natiijada labada fallaadhaha?
Jawaab:
Waxaa jira laba vector oo keliya, labada vector-na way isku toosan yihiin, sidaas darteed xalku wuxuu adeegsadaa qaacidada Pythagorean.
2. Haddii baaxadda vektorka A = 4 cutub, waxay samaysaa xagal 30 ah^o marka la isticmaalo dhidibka x ee togan, markaas baaxadda vektor-ka ee dhidibka x iyo dhidibka y waa...
Dood
Waa la ogyahay in:
A = 4 cutub, Xagal = 30^o
La weydiiyay: A_x iyo A_y ?
Jawaab:
3. Laba fallaadho xoog leh F₁ iyo F₂ mid walba oo leh cabbir dhan 5 N iyo 12 N, oo leh isla barta xiriirka isla markaana isku dhinac ah xagal dhan 60°, qiimaha ka dhashay labada fallaadhaha waa...
Dood
Waa la ogyahay in:
F₁ = 5 N, F₂ = 12 N, xagal = 60^o
Su'aal: Waa maxay natiijada labada fallaadhaha?
Jawaab:
Waxaa jira laba vector oo keliya, labada vector-na iskuma siman yihiin (waxay ku yaalliin xagal 60° ah).^o) sidaas darteed xalka dhibaatada wuxuu adeegsadaa qaacidada cosine.
4. gudaha₁ = 20 cutub iyo v₂ = 20 cutub. Waa immisa weyn tahay vector-ka natiijada keenay?

Dood
Xisaabinta vector-ka qaybaha:
v_1x =v₁ cs 30^o = (20)(½√3) = -10√3
v_1y =v₁ aan lahayn 30^o = (20)(½) = 10
v_2x =v₂ cs 30^o = (20)(½√3) = 10√3
v_2y =v₂ aan lahayn 30^o = (20)(½) = 10

v_x =v₁x + v₂x = -10√3 + 10√3 = 0
v_y =v₁y + v₂y = 10 + 10 = 20

Sharaxaad: v_1x calaamad taban sababtoo ah jihada v_1x dhanka bidix, oo ku teedsan dhidibka x ee taban. v_2x waa togan sababtoo ah jihada waxay u socotaa dhanka midig ama jihada dhidibka x ee togan. v_1y iyo v_2y waa togan sababtoo ah jihadiisu waxay kor ugu socotaa jihada dhidibka y-ga togan. Si aad u ogaato jihada vector-ka qayb kasta iyo in vector-ka qaybtu uu yahay togan ama taban, sawir vector-ka qaybtu ku yaal dhidibka x iyo dhidibka y sida ku cad tusaalaha su'aasha lambarka 2aad.

Xisaabinta vector-ka natiijada keenay:
Baaxadda vektor-ka natiijada keenay waa 20 cutub

5. Ilmuhu wuxuu orodaa masaafo dhan 80 mitir dhanka waqooyi, ka dibna wuxuu u leexdaa bari 80 mitir iyo koonfur 20 mitir. barokac waxa uu ilmuhu sameeyay waxay ahayd….

A. 60 mitir
B. 80 mitir
C. 100 mitir
D. 120 mitir
E. 180 mitir

Dood

Vector-ka Barakaca

Isticmaal qaacidada Pythagorean:
Waqooyi-bari
Jawaabta saxda ah waa C

6. Budi wuxuu u lugeeyaa 6 mitir dhanka bari, ka dibna 6 mitir dhanka koonfureed, iyo 2 mitir dhanka bari. Barakaca Budi ee ka yimid meeshii uu ka bilaabay waa...

A. 20 mitir
B. 14 mitir
C. 12 mitir
D. 10 mitir
E. 8 mitir

Dood

Isticmaal qaacidada Pythagorean:

Jihada Koonfur-bari
Jawaabta saxda ah waa D.

7.

Natiijada saddexda awoodood ee sawirka dhinaca ku jira waa…

A. 24 N
B. 16 N
C. 12 N
D. 10 N
E. 4 N

Dood
Waa la garanayaa :
F₁ = 20 Newton, Xagasha u dhaxaysa F₁ iyo dhidibka x = 0
F₂ = 20 Newton, Xagasha u dhaxaysa F₂ iyo dhidibka x = 60
F₃ = 24 Newton, Xagasha u dhaxaysa F₃ iyo dhidibka x = 60
La weydiiyay : Natiijada saddexda ciidan (F₁, F₂ iyo F₃)
Jawab :

Qaybaha fektarka xoogga ee ku yaal dhidibyada x iyo y
F_1x = (F)₁)(cos 0) = (20)(1) = 20. Togan sababtoo ah waxay ku jirtaa jihada la mid ah dhidibka x togan (midigta)
F_1y = (F)₁)(dembi 0) = (20)(0) = 0
F_2x = (F)₂)(cos 60) = (20)(0,5) = -10. Taban sababtoo ah waxay ku jirtaa jihada la mid ah dhidibka x taban (bidixda)
F_2y = (F)₂)(dembi 60) = (20)(0,5√3) = 10√3. Togan sababtoo ah waxay ku jirtaa jihada la mid ah dhidibka y togan (kor u kac)
F_3x = (F)₃)(cos 60) = (24)(0,5) = -12. Taban sababtoo ah waxay ku jirtaa jihada la mid ah dhidibka x taban (bidixda)
F_3y = (F)₃)(dembi 60) = (24)(0,5√3) = -12√3. Taban sababtoo ah waxay ku socotaa jihada la mid ah dhidibka y taban (hoos)
Natiijada qaybaha vector-ka xoogga leh ee ku yaal dhidibyada x iyo y
F_x =F_1x - F_2x - F_3x = 20 – 10 – 12 = -2
F_y =F_1y +F_2y - F_3y = 0 + 10√3 – 12√3 = -2√3
Natiijada saddexaad Tirada vektorka gaya
Jawaabta saxda ah waa E.

8.

Vektorka xoogga F₁, F₂, iyo F₃ oo ku yaal jaantuska Cartesian sida ku cad sawirka:
Natiijada saddexda vector waa…

A. √26 N
B. √76 N
C. √84 N
D. √168 N
E. √204 N

Dood
Waa la garanayaa :
F₁ = 12 Newton, Xagasha u dhaxaysa F₁ iyo dhidibka x = 30
F₂ = 10 Newton, Xagasha u dhaxaysa F₂ iyo dhidibka x = 90
F₃ = 8 Newton, Xagasha u dhaxaysa F₃ iyo dhidibka x = 30
La weydiiyay : Natiijada saddexda fallaadhaha xoogga (F₁, F₂ iyo F₃)
Jawab :
Qaybaha fektarka xoogga ee ku yaal dhidibyada x iyo y
F_1x = (F)₁)(cos 30) = (12)(0,5√3) = 6√3. Togan sababtoo ah waxay ku jirtaa jihada la mid ah dhidibka x togan (midigta)
F_1y = (F)₁)(dembiga 30) = (12)(0,5) = 6. Togan sababtoo ah waxay ku socotaa jihada la mid ah dhidibka y togan (kor u kac)
F_2x = (F)₂)(cos 90) = (10)(0) = 0.
F_2y = (F)₂)(dembi 90) = (10)(1) = -10. Taban sababtoo ah waxay ku socotaa jihada la mid ah dhidibka y taban (hoos)
F_3x = (F)₃)(cos 30) = (8)(0,5√3) = -4√3. Taban sababtoo ah waxay ku jirtaa jihada la mid ah dhidibka x taban (bidixda)
F_3y = (F)₃)(dembi 30) = (8)(0,5) = -4. Taban sababtoo ah waxay ku socotaa jihada la mid ah dhidibka y taban (hoos)
Natiijada qaybaha vector-ka xoogga leh ee ku yaal dhidibyada x iyo y
F_x =F_1x +F_2x - F_3x = 6√3 + 0 – 4√3 = 2√3
F_y =F_1y - F_2y - F_3y = 6 – 10 – 4 = -8
Natiijada saddexda fallaadhaha xoogga
Jawaabta saxda ah waa B.

9. Fiiri sawirka dhinaca. Cabbirka natiijada saddexda awoodood waa…
A. 0
B. 2√3 N
C. 4√3 N
D. 8√3 N
E. 12√3 N
Dood
Waa la garanayaa :
F₁ = 4 Newton, Xagasha u dhaxaysa F₁ iyo dhidibka x = 30
F₂ = 6√3 Newton, Xagasha u dhaxaysa F₂ iyo dhidibka x = 0
F₃ = 2 Newton, Xagasha u dhaxaysa F₃ iyo dhidibka x = 90
La weydiiyay Baaxadda natiijada saddexda ciidan (F)₁, F₂ iyo F₃)
Jawab :
Qaybaha fektarka xoogga ee ku yaal dhidibyada x iyo y
F_1x = (F)₁)(cos 30) = (4)(0,5√3) = 2√3. Togan sababtoo ah waxay ku jirtaa jihada la mid ah dhidibka x togan (midigta)
F_1y = (F)₁)(dembiga 30) = (4)(0,5) = 2. Togan sababtoo ah waxay ku socotaa jihada la mid ah dhidibka y togan (kor u kac)
F_2x = (F)₂)(cos 0) = (6√3)(1) = -6√3. Taban sababtoo ah waxay ku socotaa jihada la mid ah dhidibka x taban (bidixda)
F_2y = (F)₂)(dembi 0) = (6√3)(0) = 0.
F_3x = (F)₃)(cos 90) = (2)(0) = 0.
F_3y = (F)₃)(dembi 90) = (2)(1) = -2. Taban sababtoo ah waxay ku socotaa jihada la mid ah dhidibka y taban (hoos)
Natiijada qaybaha vector-ka xoogga leh ee ku yaal dhidibyada x iyo y
F_x =F_1x - F_2x +F_3x = 2√3 – 6√3 + 0 = -4√3
F_y =F_1y +F_2y - F_3y = 2 + 0 – 2 = 0
Natiijada saddexda fallaadhaha xoogga
Jawaabta saxda ah waa C.

10. Ilmo ayaa si toos ah u socda 10 mitir dhanka galbeed, ka dibna wuxuu u leexanayaa koonfur 4 mitir, ka dibna wuxuu u leexanayaa bari 13 mitir. Barkaca ilmuhu ka soo baxay meeshii ugu horreysay waa...
A. 4 mitir koonfur-galbeed
B. 5 mitir koonfur
C. 5 mitir dhanka koonfur-bari
D. 10 mitir bari
E. 10 mitir koonfur-bari

Dood

Jawaabta saxda ah waa C.

11. Natiijada saddexda awoodood ee sawirka dhinaca ku jira waa…

A. 1,0 N
B. 1,5 N
C. 1,9 N
D. 2,0 N
E. 2,3 N
Dood
Xisaabi baaxadda vector-ka qayb kasta:
F_1x = 10N
F_1y = 0
F_2x = -10 cos 60 = - (10) (0,5) = - 5 N
F_2y = 10 dembi 60 = (10) (0,87) = 8,7 N
F_3x = -12 cos 60 = - (12) (0,5) = - 6 N
F_3y = -12 dembi 60 = - (12) (0,87) = - 10,4 N
Go'aami vektor-ka natiijada ka soo baxday:

12. Saddex fallaadho oo xiiso leh ayaa lagu muujiyay sawirka. Cabbirka fallaadho kasta waa:
|V₁ | = 30 unug
|V₂ | = 30 unug
|V₃ | = 40 unug
Baaxadda natiijada saddexda vector waa...
A. 30 cutub
B. 40 cutub
C. 50 cutub
D. 90 cutub
E. 110 cutub
Dood
Waa la ogyahay in:
V₁ = 30, Xagasha u dhaxaysa V₁ iyo dhidibka x = 30^o
V₂ = 30, Xagasha u dhaxaysa V₂ iyo dhidibka x = 30^o
V₃ = 40, Xagasha u dhaxaysa V₃ iyo dhidibka x = 0^o
Su'aal: Natiijada saddexda fallaadhaha (V)₁, V₂ iyo V₃)
Jawaab:
Qaybaha fektarka xoogga ee ku yaal dhidibyada x iyo y
V_1x = (V₁) (koos 30^o) = (30)(0,5√3) = 15√3. Togan sababtoo ah waxay ku socotaa jihada la mid ah dhidibka x-ka togan (midigta)
V_1y = (V₁) (dembiga 30^o) = (30)(0,5) = 15. Togan sababtoo ah waxay ku jirtaa jihada la mid ah dhidibka y-ga togan (kor)
V_2x = (V₂) (koos 30^o) = (30)(0,5√3) = -15√3. Taban sababtoo ah waxay ku socotaa jihada taban ee x-dhidibka (bidixda)
V_2y = (V₂) (dembiga 30^o) = (30)(0,5) = 15. Togan sababtoo ah waxay ku jirtaa jihada la mid ah dhidibka y-ga togan (kor)
V_3x = (V₃) (koos 0^o) = (40)(1) = 40. Togan sababtoo ah waxay ku socotaa jihada la mid ah dhidibka x-ka togan (midigta)
V_3y = (V₃) (dembiga 0^o) = (40)(0) = 0

13. Laba awoodood (meel taabasho ah) ayaa is dul saaran, cabbirradooduna waa 12 N iyo 5 N. Cabbirka natiijada labada ciidan waa…
A. 17 N
B. 15 N
C. 13 N
D. 9 N
E. 7 N
Dood
Waa la ogyahay in:
Qaabka 1 (F)₁) = 12 Newton
Qaabka 2 (F)₂) = 5 Newton
La Raadinayo: Natiijada labada awoodood (ΣF)
Jawaab:
Labada awoodoodba way isku toosan yihiin sidaa darteed xoogga ka dhasha waxaa lagu xisaabiyaa qaacidada Pythagorean.

14. Saddex fallaadho oo xiiso leh ayaa lagu muujiyay sawirka soo socda. Cabbirka fallaadho kasta waa:
|V₁| = 30 unug
|V₂| = 30 unug
|V₃| = 40 unug
Baaxadda natiijada saddexda vector waa...
A. 30 cutub
B. 40 cutub
C. 50 cutub
D. 90 cutub
E. 110 cutub
Dood
Waa la ogyahay in:
v₁ = 30 cutub, oo samaynaya xagal 30 ah^o ilaa dhidibka x-ta taban.
v₂ = 30 cutub, oo samaynaya xagal 30 ah^o ilaa dhidibka x-ga ee togan.
v₃ = 40 cutub, oo samaynaya xagal 0 ah^o ilaa dhidibka x-ga ee togan.
La weydiiyay: Vector-ka natiijada leh
Jawaab:
Xisaabi qaybaha vektor-ka:
v_1x =v₁ cs 30^o = (30)(0,5√3) = -15√3 (calaamad taban sababtoo ah waxay ku socotaa jihada la mid ah dhidibka x ee taban)
v_1y =v₁ aan lahayn 30^o = (30)(0,5) = 15 (calaamad togan sababtoo ah waxay ku socotaa jihada la mid ah dhidibka y ee togan)
v_2x =v₂ cs 30^o = (30)(0,5√3) = 15√3 (calaamad togan sababtoo ah waxay ku socotaa jihada la mid ah dhidibka x-togan)
v_2y =v₂ aan lahayn 30^o = (30)(0,5) = 15 (calaamad togan sababtoo ah waxay ku socotaa jihada la mid ah dhidibka y ee togan)
v_3x =v₃ cs 0^o = (40)(1) = 40 (calaamad togan sababtoo ah waxay ku socotaa jihada la mid ah dhidibka x-togan)
v_3y =v₃ aan lahayn 0^o = (40)(0) = 0

15. Natiijada saddexda awoodood ee sawirka hoose ku jira waa...
A. 0 N
B. 2 N
C. 2√3 N
D. 3 N
E. 3√3 N
Dood
Waa la ogyahay in:
F₁ = 3 Newtons waxay sameeyaan xagal 60 ah^o dhidibka x-togan
F₂ = 3 Newtons waxay sameeyaan xagal 0 ah^o dhidibka x-ta taban
F₃ = 6 Newtons waxay sameeyaan xagal 60 ah^o dhidibka y ee taban
La weydiiyay: Xoogga natiijada
Jawaab:
Xisaabi vektor-ka qaybta:
F_1x =F₁ cs 60^o = (3)(0,5) = 1,5 N (calaamad togan sababtoo ah waxay ku socotaa jihada dhidibka x ee togan)
F_1y =F₁ aan lahayn 60^o = (3)(0,5√3) = 1,5√3 N (calaamad togan sababtoo ah waxay ku jirtaa jihada dhidibka y ee togan)
F_2x =F₂ cs 0^o = (3)(1) = -3 N (calaamad taban sababtoo ah waxay ku jirtaa jihada dhidibka x ee taban)
F_2y =F₂ aan lahayn 0^o = (3)(0) = 0
F_3x =F₃ cs 60^o = (6)(0,5) = 3 N (calaamad togan sababtoo ah waxay ku socotaa jihada dhidibka x ee togan)
F_3y =F₃ aan lahayn 60^o = (6)(0,5√3) = -3√3 N (calaamad taban sababtoo ah waxay ku jirtaa jihada dhidibka y taban)

16. Laba fallaadho xoog leh F₁ iyo F₂ mid walba oo leh cabbir dhan 15 N iyo 9 N, oo leh isla barta xiriirka isla markaana isku dhinac ah xagal dhan 60°, qiimaha ka dhashay labada fallaadhaha waa...
A. 15 N
B. 20 N
C. 21 N
D. 24 N
E. 30 N
Dood
Waa la ogyahay in:
Qaabka 1 (F)₁) = 15 Newton
Qaabka 2 (F)₂) = 9 Newton
Xagasha (θ) = 60^o
Su'aal: Natiijada labada vectors
Jawaab:
Waxaa jira laba vector oo sameeya xagal 60 ah^o si vektor-ka natiijada loo xisaabiyo iyadoo la adeegsanayo qaacidada cosine:

17. Natiijada saddexda awoodood ee sawirka dhinaca ku jira waa…
A. 24 N
B. 16 N
C. 12 N
D. 10 N
E. 4 N
Dood
Waa la ogyahay in:
F₁ = 20 Newton, Xagasha u dhaxaysa F₁ iyo dhidibka x = 0
F₂ = 20 Newton, Xagasha u dhaxaysa F₂ iyo dhidibka x = 60
F₃ = 24 Newton, Xagasha u dhaxaysa F₃ iyo dhidibka x = 60
Su'aal: Natiijada saddexda ciidan (F₁, F₂ iyo F₃)
Jawaab:
Qaybaha fektarka xoogga ee ku yaal dhidibyada x iyo y
F_1x = (F)₁)(cos 0) = (20)(1) = 20. Togan sababtoo ah waxay ku socotaa jihada la mid ah dhidibka x-ka togan (midigta)
F_1y = (F)₁)(dembi 0) = (20)(0) = 0
F_2x = (F)₂)(cos 60) = (20)(0,5) = -10. Taban sababtoo ah waxay ku socotaa jihada taban ee x-dhidibka (bidixda)
F_2y = (F)₂)(dembi 60) = (20)(0,5√3) = 10√3. Togan sababtoo ah waxay ku socotaa jihada la mid ah dhidibka y-ga togan (kor)
F_3x = (F)₃)(cos 60) = (24)(0,5) = -12. Taban sababtoo ah waxay ku socotaa jihada taban ee x-dhidibka (bidixda)
F_3y = (F)₃)(dembi 60) = (24)(0,5√3) = -12√3. Taban sababtoo ah waxay ku socotaa jihada taban ee dhidibka y (hoos u socda)

18. Shay ayaa ka dhaqaaqa E una gudba F wuxuuna ku dhammaadaa G. Sawirka hoose, oo muujinaya barokac 10 cutub ah, waa...

Dood

Isku-duwayaasha dhibcaha E = x, y = 1, 1

Isku-duwayaasha dhibcaha F = x, y = 9, 1

Isku-duwayaasha dhibcaha G = x, y = 9, 7

Dhererka EF = 9-1 = 8

Dhererka FG = 7-1 = 6

Dhererka EG =

Jawaabta saxda ah waa A.

19. Qofku wuxuu gaari ku safraa A ilaa B, 30 km waqooyi, ka dibna wuxuu u sii socdaa C, 60 km bari, ugu dambayntiina wuxuu gaaraa magaalada D, 110 km koonfur. Baxsiga gaariga ee A ilaa D waa...

A. 200 km

B. 140 km

Q. 120 km

D. 100 km

Bariga 80 km

Dood

AA' = 60 km

A'D = 110 km - 30 km = 80 km

Jawaabta saxda ah waa D.

20. Andi wuxuu gaari uga safrayaa magaalada A waqooyi ilaa magaalada B 100 km, ka dibna wuxuu sii waday safarkiisii ​​​​magaalada C bari 60 km, ka dibna Andi wuxuu u safray koonfurta ilaa magaalada D 20 km. Barkaca gaariga waa…

A. 10 km

B. 20 km

Q. 80 km

D. 100 km

Bariga 180 km

Dood

D'D = 60 km

AD' = 100 km - 20 km = 80 km

Jawaabta saxda ah waa D.

21. Munaasabaddii "City Marathon Festival" ee Oktoobar 2014 ee Jakarta waxaa ka dhacay 4 qaybood oo orod ah, kuwaas oo kala ah qaybta marathon buuxa (42 km), qaybta nus maarad ah (21 km), qaybta 10 kiiloomitir iyo qaybta 5 kiiloomitir halkaas oo lagu go'aamiyay tartanka qayb kasta. Tartankan marathonka ah wuxuu ka bilaabmayaa Xarunta Ciyaaraha Bung Karno wuxuuna ku dhammaadaa Taallada Qaranka (Monas). Mid ka mid ah ka qaybgalayaasha tartanka, Andri, ayaa ka qayb qaatay tartanka. marathon buuxa wuxuuna awoodaa inuu ka safro wadada kaliya meelaha A, B, iyo C sida Jaantuska 2aad.

Haddii 1 sanduuq uu matalo 1 km, markaa wadarta guud ee barakaca uu Andri sameeyay waa…

A. 26 km

B. 20 km

Q. 12 km

D. 10 km

Bariga 8 km

Dood

Dhererka dhinaca hoose = 8 km, dhererka dhinaca hore = 6 km.

Iyadoo la adeegsanayo qaacidada Pythagorean, Barakaca = R = 10 km

Jawaabta saxda ah waa D.

22. Qofku wuxuu gaari ku safraa A ilaa B, 30 km waqooyi, ka dibna wuxuu u sii socdaa C, 60 km bari, ugu dambayntiina wuxuu gaaraa magaalada D, 110 km koonfur. Baxsiga gaariga ee A ilaa D waa...

A. 200 km

B. 140 km

Q. 120 km

D. 100 km

Bariga 80 km

Dood

AA' = 60 km

A'D = 110 km - 30 km = 80 km

Jawaabta saxda ah waa D.

23. Andi wuxuu gaari uga safrayaa magaalada A waqooyi ilaa magaalada B 100 km, ka dibna wuxuu sii waday safarkiisii ​​​​magaalada C bari 60 km, ka dibna Andi wuxuu u safray koonfurta ilaa magaalada D 20 km. Barkaca gaariga waa…

A. 10 km

B. 20 km

Q. 80 km

D. 100 km

Bariga 180 km

Dood

D'D = 60 km

AD' = 100 km - 20 km = 80 km

Jawaabta saxda ah waa D.

Vektor-ka Cutubka

24. Su'aalaha Imtixaanka Qaranka ee 2000/2001
Shay asal ahaan ku sugnaa barta tixraaca ayaa ku dhaqaaqa xawaare v = (2i − 1,5h) ms^-1Ka dib markii uu dhaqaaqay 4 ilbiriqsi, shaygu wuxuu u dhaqaaqay masaafo dhan…
A. 2 mitir
B. 10 mitir
C. 12 mitir
D. 14 mitir
E. 25 mitir
Dood
Waa la ogyahay in:
Xawaaraha jihada toosan (v_x) = 2 m/s
Xawaaraha jihada toosan (v_y) = 1,5 m/s
Muddada (t) = 4 ilbiriqsi
La weydiiyay: Barakaca walxaha
Jawaab:
Xisaabi xawaaraha ka dhashay shayga (v):

25. Su'aalaha Imtixaanka Qaranka ee 2007/2008 P4 No. 3
Vektor F₁ = 14 N iyo F₂ = 10 N waxaa lagu dhejiyay jaantuska Cartesian sida ku cad sawirka. Vektorka natiijada leh haddii lagu muujiyo vektorrada cutubka R = i + j waa….
A. 7i + 10√3 j
B. 7i + 10j
C. 3i + 7√3 j
D. 3i + 10j
E. 3i + 7j
Dood
Xisaabinta vector-ka qaybaha:
F_1x = (F)₁) (koos 60^o) = (14)(0,5) = -7 N (calaamad taban sababtoo ah waxay ku jirtaa jihada taban ee x)
F_1y = (F)₁) (dembiga 60^o) = (14)(0,5√3) = 7√3 N (calaamad togan sababtoo ah waxay ku socotaa jihada y ee togan)
F_2x = 10N
F_2y = 0
Xisaabinta vector-ka qaybaha ka dhashay:
F_x =F_1x +F_2x +F_3x = -7 + 10 = 3 N
F_y =F_1y +F_2y +F_3y = 7√3 + 0 = 7√3 N
Vektorka natiijada leh haddii lagu muujiyo vektorrada cutubka:
R = 3 i + 7√3 j
Jawaabta saxda ah waa C.

 

Isha su'aasha:

Su'aalaha Imtixaanka Qaranka ee Fiisigiska ee Dugsiga Sare/Dugsiga Sare ee Xirfadda

Su'aalaha Vektor-ka
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