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Berikut ini adalah pertanyaan dari pakaibaru123 pada mata pelajaran Fisika untuk jenjang Sekolah Menengah Atas

Bahasa Inggris x fisika gaes help ya

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Use graph paper to determine the magnitude of displacement of a car that travels :

A. 10 km due south, 2 km due west, then 5 km due east.

B. 10 km due east and then 12 km at 45° north of west.

A. The magnitude of displacement is 10.44 km.
The direction of displacement is 286.7° from x axes positive.

B. The magnitude of displacement is 20.35 km.
The direction of displacement is 24.66° from x axes positive.

Penjelasan dengan langkah-langkah:

Diketahui:

A. d₁ = 10 km
θ₁ = 270° from x axes positive
d₂ = 2 km
θ₂ = 180°
d₃ = 5 km
θ₃ = 0°
B. d₁ = 10 km
θ₁ = 0° from x axes positive
d₂ = 12 km
θ₂ = 45°

Ditanyakan:

Magnitude?
Direction?

Jawaban:

A. Look each distance in x-axes and y-axes.

d₁ = 10 km and θ₁ = 270°
$d_{1x} \:=\: d_1 \: cos \: 270^o \:=\: 10 \times 0$ = 0 km
$d_{1y} \:=\: d_1 \: sin \: 270^o \:=\: 10 \times - 1$ = - 10 km
d₂ = 2 km and θ₂ = 180°
$d_{2x} \:=\: d_2 \: cos \: 180^o \:=\: 2 \times - 1$ = - 2 km
$d_{2y} \:=\: d_2 \: sin \: 180^o \:=\: 2 \times 0$ = 0 km
d₃ = 5 km and θ₃ = 0°
$d_{3x} \:=\: d_3 \: cos \: 0^o \:=\: 5 \times 1$ = 5 km
$d_{3y} \:=\: d_3 \: sin \: 0^o \:=\: 5 \times 0$ = 0 km

Displacement in x-axes and y-axes.

$d_x \:=\: d_{1x} \:+\: d_{2x} \:+\: d_{3x}$
$d_x \:=\: 0 \:+\: (- 2) \:+\: 5$
dₓ = 3 km
$d_y \:=\: d_{1y} \:+\: d_{2y} \:+\: d_{3y}$
$d_y \:=\: (- 10) \:+\: 0 \:+\: 0$
dy = - 10 km

Magnitude of displacement

$d^2 \:=\: d_x^2 \:+\: d_y^2$
$d^2 \:=\: 3^2 \:+\: (- 10)^2$
$d^2 \:=\: 9 \:+\: 100$
$d^2 \:=\: 109$
$d \:=\: \sqrt{109}$
d = 10.44 km
Direction
$tan \: \theta \:=\: \frac{d_y}{d_x}$
$tan \: \theta \:=\: \frac{- 10}{3}$
$\theta \:=\: tan^{- 1} \: - 3.33$
$\theta \:=\: - 73.3^o$
$\theta \:=\: 360^o \:-\: 73.3^o$
θ = 286.7°

B. Look each distance in x-axes and y-axes.

d₁ = 10 km and θ₁ = 0°
$d_{1x} \:=\: d_1 \: cos \: 0^o \:=\: 10 \times 1$ = 10 km
$d_{1y} \:=\: d_1 \: sin \: 0^o \:=\: 10 \times 0$ = 0 km
d₂ = 12 km and θ₂ = 45°
$d_{2x} \:=\: d_2 \: cos \: 45^o \:=\: 12 \times \frac{1}{2} \sqrt{2}$ = 8.49 km
$d_{2y} \:=\: d_2 \: sin \: 45^o \:=\: 12 \times \frac{1}{2} \sqrt{2}$ = 8.49 km

Displacement in x-axes and y-axes.

$d_x \:=\: d_{1x} \:+\: d_{2x}$
$d_x \:=\: 10 \:+\: 8.49$
dₓ = 18.49 km
$d_y \:=\: d_{1y} \:+\: d_{2y}$
$d_y \:=\: 0 \:+\: 8.49$
dy = 8.49 km

Magnitude of displacement

$d^2 \:=\: d_x^2 \:+\: d_y^2$
$d^2 \:=\: 18.49^2 \:+\: 8.49^2$
$d^2 \:=\: 341.8801 \:+\: 72.0801$
$d^2 \:=\: 413.9602$
$d \:=\: \sqrt{413.9602}$
d = 20.35 km
Direction
$tan \: \theta \:=\: \frac{d_y}{d_x}$
$tan \: \theta \:=\: \frac{8.49}{18.49}$
$\theta \:=\: tan^{- 1} \: 0.459$
θ = 24.66°

Pelajari lebih lanjut

Pelajari lebih lanjut tentang Resultan Tiga Vektor yomemimo.com/tugas/12527736

#BelajarBersamaBrainly

#SPJ1

$Use graph paper to determine the magnitude of displacement of a car that travels : A. 10 km due south, 2 km due west, then 5 km due east.B. 10 km due east and then 12 km at 45° north of west.A. The magnitude of displacement is 10.44 km.The direction of displacement is 286.7° from x axes positive.B. The magnitude of displacement is 20.35 km.The direction of displacement is 24.66° from x axes positive.Penjelasan dengan langkah-langkah:Diketahui:A. d₁ = 10 kmθ₁ = 270° from x axes positived₂ = 2 km θ₂ = 180°d₃ = 5 kmθ₃ = 0°B. d₁ = 10 kmθ₁ = 0° from x axes positived₂ = 12 km θ₂ = 45°Ditanyakan:Magnitude?Direction?Jawaban:A. Look each distance in x-axes and y-axes.d₁ = 10 km and θ₁ = 270°[tex]d_{1x} \:=\: d_1 \: cos \: 270^o \:=\: 10 \times 0[/tex] = 0 km[tex]d_{1y} \:=\: d_1 \: sin \: 270^o \:=\: 10 \times - 1[/tex] = - 10 kmd₂ = 2 km and θ₂ = 180°[tex]d_{2x} \:=\: d_2 \: cos \: 180^o \:=\: 2 \times - 1[/tex] = - 2 km[tex]d_{2y} \:=\: d_2 \: sin \: 180^o \:=\: 2 \times 0[/tex] = 0 kmd₃ = 5 km and θ₃ = 0°[tex]d_{3x} \:=\: d_3 \: cos \: 0^o \:=\: 5 \times 1[/tex] = 5 km[tex]d_{3y} \:=\: d_3 \: sin \: 0^o \:=\: 5 \times 0[/tex] = 0 kmDisplacement in x-axes and y-axes.[tex]d_x \:=\: d_{1x} \:+\: d_{2x} \:+\: d_{3x}[/tex][tex]d_x \:=\: 0 \:+\: (- 2) \:+\: 5[/tex]dₓ = 3 km[tex]d_y \:=\: d_{1y} \:+\: d_{2y} \:+\: d_{3y}[/tex][tex]d_y \:=\: (- 10) \:+\: 0 \:+\: 0[/tex]dy = - 10 kmMagnitude of displacement[tex]d^2 \:=\: d_x^2 \:+\: d_y^2[/tex][tex]d^2 \:=\: 3^2 \:+\: (- 10)^2[/tex][tex]d^2 \:=\: 9 \:+\: 100[/tex][tex]d^2 \:=\: 109[/tex][tex]d \:=\: \sqrt{109}[/tex]d = 10.44 kmDirection[tex]tan \: \theta \:=\: \frac{d_y}{d_x}[/tex][tex]tan \: \theta \:=\: \frac{- 10}{3}[/tex][tex]\theta \:=\: tan^{- 1} \: - 3.33[/tex][tex]\theta \:=\: - 73.3^o[/tex][tex]\theta \:=\: 360^o \:-\: 73.3^o[/tex]θ = 286.7°B. Look each distance in x-axes and y-axes.d₁ = 10 km and θ₁ = 0°[tex]d_{1x} \:=\: d_1 \: cos \: 0^o \:=\: 10 \times 1[/tex] = 10 km[tex]d_{1y} \:=\: d_1 \: sin \: 0^o \:=\: 10 \times 0[/tex] = 0 kmd₂ = 12 km and θ₂ = 45°[tex]d_{2x} \:=\: d_2 \: cos \: 45^o \:=\: 12 \times \frac{1}{2} \sqrt{2}[/tex] = 8.49 km[tex]d_{2y} \:=\: d_2 \: sin \: 45^o \:=\: 12 \times \frac{1}{2} \sqrt{2}[/tex] = 8.49 kmDisplacement in x-axes and y-axes.[tex]d_x \:=\: d_{1x} \:+\: d_{2x}[/tex][tex]d_x \:=\: 10 \:+\: 8.49[/tex]dₓ = 18.49 km[tex]d_y \:=\: d_{1y} \:+\: d_{2y}[/tex][tex]d_y \:=\: 0 \:+\: 8.49[/tex]dy = 8.49 kmMagnitude of displacement[tex]d^2 \:=\: d_x^2 \:+\: d_y^2[/tex][tex]d^2 \:=\: 18.49^2 \:+\: 8.49^2[/tex][tex]d^2 \:=\: 341.8801 \:+\: 72.0801[/tex][tex]d^2 \:=\: 413.9602[/tex][tex]d \:=\: \sqrt{413.9602}[/tex]d = 20.35 kmDirection[tex]tan \: \theta \:=\: \frac{d_y}{d_x}[/tex][tex]tan \: \theta \:=\: \frac{8.49}{18.49}[/tex][tex]\theta \:=\: tan^{- 1} \: 0.459[/tex]θ = 24.66°Pelajari lebih lanjutPelajari lebih lanjut tentang Resultan Tiga Vektor https://brainly.co.id/tugas/12527736#BelajarBersamaBrainly#SPJ1$

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Last Update: Sun, 27 Nov 22

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