Roboguru

Tiga buah vektor sebagai berikut: = 3i + 5j + 2k = 5i – 3k = 6j + 2k Hasil dari adalah ….

Pertanyaan

Tiga buah vektor sebagai berikut:
L with rightwards harpoon with barb upwards on top = 3i + 5j + 2k
M with rightwards harpoon with barb upwards on top = 5i – 3k
N with rightwards harpoon with barb upwards on top= 6j + 2k
Hasil dari L with rightwards harpoon with barb upwards on top minus 2 M with rightwards harpoon with barb upwards on top plus 3 N with rightwards harpoon with barb upwards on top adalah ….

  1. 7i + 23j + 14k  

  2. 7i + 23j - 14k  

  3. -7i + 23j + 14k   

  4. -7i - 23j + 14k  

  5. 7i - 23j - 14k   

Pembahasan Video:

Pembahasan Soal:

Dit : L with rightwards harpoon with barb upwards on top minus 2 M with rightwards harpoon with barb upwards on top plus 3 N with rightwards harpoon with barb upwards on top ?

Dik:

L with rightwards harpoon with barb upwards on top = 3i + 5j + 2k
M with rightwards harpoon with barb upwards on top = 5i – 3k
N with rightwards harpoon with barb upwards on top= 6j + 2k

menggunakan konsep pengurangan dan pertambahan dalam vektor :

equals left parenthesis 3 straight i space plus space 5 straight j space plus space 2 straight k space right parenthesis minus 2 left parenthesis 5 straight i space – space 3 straight k right parenthesis plus 3 left parenthesis space 6 straight j space plus space 2 straight k right parenthesis equals negative 7 straight i plus 23 straight j plus 14 straight k

Jadi, jawaban yang tepat adalah C

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 30 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Vektor gaya , , dan  terletak pada sebuah diagram kartesius seperti gambar : Resultan ketiga vektor adalah ...

Pembahasan Soal:

Diketahui :

straight F subscript 1 equals 12 space straight N  straight F subscript 2 equals 10 space straight N  straight F subscript 3 equals 8 space straight N  θF subscript 1 equals 30 degree  θF subscript 3 equals 90 degree minus 60 degree equals 30 degree  straight F subscript straight R ?

Tinjau sumbu x :

sum straight F subscript straight x equals straight F subscript 1 space cos space 30 degree space minus space straight F subscript 3 space cos space 30 degree  sum straight F subscript straight x equals 12 space open parentheses 1 half square root of 3 close parentheses space minus space 8 open parentheses 1 half square root of 3 close parentheses  sum straight F subscript straight x equals 6 square root of 3 minus 4 square root of 3  sum straight F subscript straight x equals 2 square root of 3 space straight N

Tinjau sumbu y :

sum straight F subscript straight y equals straight F subscript 1 space sin space 30 degree space minus space straight F subscript 3 space sin space 30 degree minus space straight F subscript 2  sum straight F subscript straight y equals 12 space open parentheses 1 half close parentheses space minus space 8 open parentheses 1 half close parentheses minus 10  sum straight F subscript straight y equals 6 minus 4 minus 10  sum straight F subscript straight y equals negative 8 space straight N

Maka, besarnya resultan gaya dengan menggunakan vektor adalah :

straight F subscript straight R equals square root of open parentheses sum straight F subscript straight x close parentheses squared plus space open parentheses sum straight F subscript straight y close parentheses squared end root  straight F subscript straight R equals square root of open parentheses 2 square root of 3 close parentheses squared plus space open parentheses negative 8 close parentheses squared end root  straight F subscript straight R equals square root of 76 space straight N

1

Roboguru

Perhatikan gambar vektor dibawah ini!   Jika besar vektor  adalah 24 N dan besar vektor  adalah 10 N, besar resultan kedua vektor tersebut adalah ....

Pembahasan Soal:

Uraikan vektor-vektor tersebut dalam arah vertikal dan horizontal seperti berikut ini.

Jumlah komponen vektor dalam arah horizontal (sumbu x) adalah:

begin mathsize 14px style sum for space of begin inline style F subscript x end style equals F subscript 2 space sin space 37 to the power of straight o begin inline style sum for space of end style begin inline style F subscript x end style begin inline style equals end style begin inline style 10 end style begin inline style open parentheses begin display style 0 comma 6 end style close parentheses end style begin inline style sum for space of end style begin inline style F subscript x end style begin inline style equals end style 6 space begin inline style straight N end style end style  

Jumlah komponen vektor dalam arah vertikal (sumbu y) adalah:

begin mathsize 14px style sum for space of F subscript y equals F subscript 1 space minus F subscript 2 space cos space 37 to the power of straight o begin inline style sum for space of end style begin inline style F subscript y end style begin inline style equals end style 24 minus 10 left parenthesis 0 comma 8 right parenthesis begin inline style sum for space of end style begin inline style F subscript y end style begin inline style equals end style 24 minus 8 begin inline style sum for space of end style begin inline style F subscript y end style begin inline style equals end style 16 space begin inline style straight N end style end style 

Maka besar resultan kedua vektor dapat dicari dengan penguraian vektor.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line R with rightwards arrow on top vertical line end cell equals cell square root of left parenthesis sum open vertical bar stack F subscript x with rightwards arrow on top close vertical bar right parenthesis squared plus left parenthesis sum open vertical bar stack F subscript y with rightwards arrow on top close vertical bar right parenthesis squared end root end cell row cell vertical line R with rightwards arrow on top vertical line end cell equals cell square root of left parenthesis 6 right parenthesis squared plus left parenthesis 16 right parenthesis squared end root end cell row cell vertical line R with rightwards arrow on top vertical line end cell equals cell square root of 292 end cell row cell vertical line R with rightwards arrow on top vertical line end cell equals cell 2 square root of 73 blank straight N end cell end table end style  

Jadi, jawaban yang tepat adalah A.    

0

Roboguru

Jika besarnya resultan dan selisih kedua vektor sama, maka besarnya sudut apit kedua vektor adalah ….

Pembahasan Soal:

Diketahui: Resultan vektor = Selisih vektor

Ditanya: sudut apit (θ) = ...?

Jawaban:

Resultan vektor dirumuskan sebagai:

R equals square root of open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared plus 2 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar space cos space theta end root 

Sedangkan selisih vektor dirumuskan sebagai:

S equals square root of open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared minus 2 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar space cos space theta end root  

Maka, 

table attributes columnalign right center left columnspacing 0px end attributes row R equals S row cell square root of open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared plus 2 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar space cos space theta end root end cell equals cell square root of open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared minus 2 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar space cos italic space theta end root end cell row cell open parentheses square root of open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared plus 2 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar space cos space theta end root close parentheses squared end cell equals cell open parentheses square root of open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared minus 2 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar space cos space theta end root close parentheses squared end cell row cell open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared plus 2 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar space cos space theta end cell equals cell open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared minus 2 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar space cos space theta end cell row cell 2 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar space cos space theta plus 2 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar space cos space theta end cell equals 0 row cell 4 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar space cos italic space theta end cell equals 0 row cell cos space theta end cell equals 0 row theta equals cell cos to the power of negative 1 end exponent open parentheses 0 close parentheses end cell row theta equals cell 90 degree end cell end table   

Dengan demikian, besar sudut apit kedua vektor adalah 90°.

Jadi, tidak ada jawaban yang tepat.

1

Roboguru

Dua buah gaya yang sama besar 10 N membentuk sudut 120° satu sama lain. Selisih kedua vektor tersebut adalah ...

Pembahasan Soal:

Diketahui:

begin mathsize 14px style F subscript 1 equals F subscript 2 equals 10 space N alpha equals 120 to the power of 0 end style 

Ditanyabegin mathsize 14px style S end style ?

Resultan gaya adalah hasil operasi hitung antara dua vektor gaya atau lebih

Penyelesaian:

S equals F subscript 1 minus F subscript 2 S equals square root of F subscript 1 squared plus F subscript 2 squared minus 2 F subscript 1 F subscript 2 cos alpha end root S equals square root of 10 squared plus 10 squared minus 2.10.10 cos left parenthesis 120 right parenthesis end root S equals square root of 200 minus 200 left parenthesis negative bevelled 1 half right parenthesis end root S equals square root of 300 S equals 10 square root of 3 space end root straight N 

Jadi, selisih kedua vektor tersebut adalah begin mathsize 14px style 10 square root of 3 space straight N end style. (D)undefined 

0

Roboguru

Vektor , , dan  diletakkan pada diagram Cartesius seperti pada gambar. Berapa resultan ketiga vektor tersebut?

Pembahasan Soal:

Error converting from MathML to accessible text.

 

Hitung total gaya di sumbu - x:

Error converting from MathML to accessible text.

 

Hitung total gaya di sumbu - y:

Error converting from MathML to accessible text.

 

Maka total gayanya adalah :

begin mathsize 14px style begin inline style straight F subscript straight R equals square root of left parenthesis straight capital sigma space straight F subscript straight x right parenthesis squared plus left parenthesis straight capital sigma space straight F subscript straight y right parenthesis squared end root  straight F subscript straight R equals square root of left parenthesis 8 right parenthesis squared plus left parenthesis 6 right parenthesis squared end root  straight F subscript straight R equals square root of 100  straight F subscript straight R equals 10 space straight N end style end style

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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