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3 years ago

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A vector has a magnitude of 46.0 m and points in a direction 20.0° below the positive x-axis. A second vector, , has a magnitude

of 86.0 m and points in a direction 42.0° above the negative x-axis. a) Sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗ .
b) Using the component method of vector addition, find the magnitude of the vector C⃗ .
c) Using the component method of vector addition, find the direction of the vector C

1 answer:

irina1246 [14]3 years ago

6 0

Answer with Step-by -step explanation:

We are given that

b. $\mid A\mid=46 m$

$\theta=20^{\circ}$ below the positive x-axis

Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis= $x=360-20=340^{\circ}$

x-component of vector A= $A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24$

y-Component of vector A= $A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64$

Magnitude of vector B=86 m

The vector B makes angle with positive x- axis= $x'=42^{\circ}$

x-component of vector B= $B_x=86cos42=63.64$

y-Component of vector B= $B_y=86sin42=57.62$

Vector A= $A_xi+A_yj=43.24i-15.64j$

Vector B= $B_xi+B_yj=63.64i+57.62j$

Vector C=A+B

Substitute the values

$C=43.24i-15.64j+63.64i+57.62j$

$C=106.88i+41.98j$

c.Direction= $\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}$

The direction of the vector C=21.5 degree

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<u>Givens:</u>

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See below ↓

Explanation:

<u>Step 1 : Diagram</u>

In attachment

<u>Step 2</u>

We choose the system to be the spring, the block, and the Earth and it is isolated
We put all the data in the figure we have created and create a zero level (initial height) of the block to be yₓ = 0 and the final position, when it stops and moves upwards again, to be yₙ = -A
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<u>Step 3</u>

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<u>Step 4</u>

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<u>Step 5</u>

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<u>Step 6</u>

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8 0

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