Identifique la dirección de vector E → = ( 15 k g f , N 15 º O ) \overrightarrow{E\ \ }=\left(15\ kgf,\ N\ 15º\ O\right) E <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 1 5 k g f , N 1 5 º O )

α = 105 º \alpha=105º α = 1 0 5 º

α = 75 º \alpha=75º α = 7 5 º

α = 255 º \alpha=255º α = 2 5 5 º

α = 285 º \alpha=285º α = 2 8 5 º

Identifique las coordenadas rectangulares del vector G → = 15 N ( 0.8 i → − 0.6 j → ) \overrightarrow{G\ }=15\ N\ \left(0.8\ \overrightarrow{\ i\ }-0.6\ \overrightarrow{\ j\ }\right) G <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = 1 5 N ( 0 . 8 i <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> − 0 . 6 j <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> )

G → = ( 12 , − 9 ) \overrightarrow{G\ }=\left(12\ ,\ -9\right) G <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 1 2 , − 9 )

G → = ( 0.8 , − 0.6 ) \overrightarrow{G\ }=\left(0.8\ ,\ -0.6\right) G <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 0 . 8 , − 0 . 6 )

G → = ( 9 , 12 ) \overrightarrow{G\ }=\left(9\ ,\ 12\right) G <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 9 , 1 2 )

( − 0.6 , 0.8 ) \left(-0.6\ ,\ 0.8\right) ( − 0 . 6 , 0 . 8 )

Identifique el módulo del vector R → = ( 5 i → − 2 j → ) m \overrightarrow{R\ }=\left(5\ \overrightarrow{\ i\ }\ -2\ \overrightarrow{\ j\ }\right)\ m R <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 5 i <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> − 2 j <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ) m

∣ R → ∣ = 5.39 \left|\overrightarrow{R\ }\right|=5.39 ∣ ∣ ∣ R <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ∣ ∣ ∣ = 5 . 3 9

∣ R → ∣ = 4.58 \left|\overrightarrow{R\ }\right|=4.58 ∣ ∣ ∣ R <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ∣ ∣ ∣ = 4 . 5 8

∣ R → ∣ = 3.87 \left|\overrightarrow{R\ }\right|=3.87 ∣ ∣ ∣ R <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ∣ ∣ ∣ = 3 . 8 7

∣ R → ∣ = 2.45 \left|\overrightarrow{R\ }\right|=2.45 ∣ ∣ ∣ R <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ∣ ∣ ∣ = 2 . 4 5

Si el vector D → = ( 2 , − 8 ) \overrightarrow{D\ }=\left(2\ ,\ -8\right) D <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = ( 2 , − 8 ) tiene un ángulo θ = ∣ tan ⁡ − 1 ( − 8 2 ) ∣ = 75.96 º \theta=\left|\tan^{-1}\left(\frac{-8}{2}\right)\right|=75.96º θ = ∣ ∣ ∣ ∣ tan − 1 ( 2 − 8 ) ∣ ∣ ∣ ∣ = 7 5 . 9 6 º identifique la dirección del vector

α = 284.04 º \alpha=284.04º α = 2 8 4 . 0 4 º

α = 75.96 º \alpha=75.96º α = 7 5 . 9 6 º

α = 104.04 º \alpha=104.04º α = 1 0 4 . 0 4 º

α = 255.96 º \alpha=255.96º α = 2 5 5 . 9 6 º

VECTORES

Quiz

•

Physics

•

11th Grade

•

Practice Problem

•

Hard

Diana Montenegro

Used 310+ times

FREE Resource

10 questions

Show all answers

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Identifique el cuadrante en el que se ubica el vector

$\overrightarrow{A\ }=\left(-7,5\right)\ m$

I cuadrante

II cuadrante

III cuadrante

IV cuadrante

MULTIPLE SELECT QUESTION

45 sec • 1 pt

Para expresar un vector en coordenadas polares, escribimos el par ordenado ( , )

(módulo , dirección)

(módulo, rumbo)

(componente x , componente y)

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

A partir del gráfico, es correcto mencionar que:

a) La dirección del vector es 233.13º

b) El rumbo del vector es S 53.13º O

c) Al emplear la fórmula $\tan^{-1}\left(\frac{Py}{Px}\right)$ obtendremos como resultado un ángulo de 36.87º

Todas las anteriores

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Para escribir su rumbo de un vector que está ubicado en el IV cuadrante, se emplean los puntos cardinales

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Identifique la dirección de vector

$\overrightarrow{E\ \ }=\left(15\ kgf,\ N\ 15º\ O\right)$

$\alpha=75º$

$\alpha=105º$

$\alpha=255º$

$\alpha=285º$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Para expresar un vector en función de sus vectores base se debe escribir

$\left(componente\ x\ ,\ componente\ y\right)$

$modulo\ \left(\frac{componente\ x}{modulo}\ \overrightarrow{i\ \ }+\frac{componente\ y}{modulo}\overrightarrow{\ j\ }\right)$

$\left(modulo,\ direccion\right)$

$\left(componente\ x\ \overrightarrow{\ i\ }+componente\ y\ \overrightarrow{\ j\ }\right)$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Identifique las coordenadas rectangulares del vector
$\overrightarrow{G\ }=15\ N\ \left(0.8\ \overrightarrow{\ i\ }-0.6\ \overrightarrow{\ j\ }\right)$

$\overrightarrow{G\ }=\left(0.8\ ,\ -0.6\right)$

$\overrightarrow{G\ }=\left(12\ ,\ -9\right)$

$\overrightarrow{G\ }=\left(9\ ,\ 12\right)$

$\left(-0.6\ ,\ 0.8\right)$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Identifique el módulo del vector

$\overrightarrow{R\ }=\left(5\ \overrightarrow{\ i\ }\ -2\ \overrightarrow{\ j\ }\right)\ m$

$\left|\overrightarrow{R\ }\right|=4.58$

$\left|\overrightarrow{R\ }\right|=3.87$

$\left|\overrightarrow{R\ }\right|=5.39$

$\left|\overrightarrow{R\ }\right|=2.45$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Si el vector $\overrightarrow{D\ }=\left(2\ ,\ -8\right)$ tiene un ángulo $\theta=\left|\tan^{-1}\left(\frac{-8}{2}\right)\right|=75.96º$ identifique la dirección del vector

$\alpha=75.96º$

$\alpha=104.04º$

$\alpha=255.96º$

$\alpha=284.04º$

10.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Dado el vector $\overrightarrow{B\ }=\left(7\ N,\ 210º\right)$ , identifique el ángulo que se emplea para representar su rumbo

60º

30º

210º

80º

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VECTORES

10 questions

Identifique el cuadrante en el que se ubica el vector A →=(−7,5) m\overrightarrow{A\ }=\left(-7,5\right)\ mA =(−7,5) m

Para expresar un vector en coordenadas polares, escribimos el par ordenado ( , )

A partir del gráfico, es correcto mencionar que:

Para escribir su rumbo de un vector que está ubicado en el IV cuadrante, se emplean los puntos cardinales

Identifique la dirección de vector E →=(15 kgf, N 15º O)\overrightarrow{E\ \ }=\left(15\ kgf,\ N\ 15º\ O\right)E =(15 kgf, N 15º O)

Para expresar un vector en función de sus vectores base se debe escribir

Identifique las coordenadas rectangulares del vector G →=15 N (0.8 i →−0.6 j →)\overrightarrow{G\ }=15\ N\ \left(0.8\ \overrightarrow{\ i\ }-0.6\ \overrightarrow{\ j\ }\right)G =15 N (0.8 i −0.6 j ​)

Identifique el módulo del vector R →=(5 i → −2 j →) m\overrightarrow{R\ }=\left(5\ \overrightarrow{\ i\ }\ -2\ \overrightarrow{\ j\ }\right)\ mR =(5 i −2 j ​) m

Si el vector D →=(2 , −8)\overrightarrow{D\ }=\left(2\ ,\ -8\right)D =(2 , −8) tiene un ángulo θ=∣tan⁡−1(−82)∣=75.96º\theta=\left|\tan^{-1}\left(\frac{-8}{2}\right)\right|=75.96ºθ=∣∣∣∣​tan−1(2−8​)∣∣∣∣​=75.96º identifique la dirección del vector

Dado el vector B →=(7 N, 210º)\overrightarrow{B\ }=\left(7\ N,\ 210º\right)B =(7 N, 210º) , identifique el ángulo que se emplea para representar su rumbo

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Identifique el cuadrante en el que se ubica el vector

$\overrightarrow{A\ }=\left(-7,5\right)\ m$

Identifique la dirección de vector

$\overrightarrow{E\ \ }=\left(15\ kgf,\ N\ 15º\ O\right)$

Identifique las coordenadas rectangulares del vector
$\overrightarrow{G\ }=15\ N\ \left(0.8\ \overrightarrow{\ i\ }-0.6\ \overrightarrow{\ j\ }\right)$

Identifique el módulo del vector

$\overrightarrow{R\ }=\left(5\ \overrightarrow{\ i\ }\ -2\ \overrightarrow{\ j\ }\right)\ m$

Si el vector $\overrightarrow{D\ }=\left(2\ ,\ -8\right)$ tiene un ángulo $\theta=\left|\tan^{-1}\left(\frac{-8}{2}\right)\right|=75.96º$ identifique la dirección del vector

Dado el vector $\overrightarrow{B\ }=\left(7\ N,\ 210º\right)$ , identifique el ángulo que se emplea para representar su rumbo