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lara31 [8.8K]
3 years ago
15

Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length firs

t then the degrees.
Physics
1 answer:
mafiozo [28]3 years ago
4 0

Answer:

Magnitude of the vector is 23.75\ \text{units} and the direction is 35.23^{\circ}

Explanation:

Magnitude of first vector = |A| = 13\ \text{units}

Angle = \theta_1=27^{\circ}

Magnitude of second vector = |B| = 11\ \text{units}

Angle = \theta_2=45^{\circ}

x component of first vector

A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}

y component of first vector

A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}

x component of second vector

B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}

y component of first vector

B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}

Adding the magnitudes

C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}

C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}

Magnitude of the sum of the vectors would be

|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}

The direction would be

\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}

The magnitude of the vector is 23.75\ \text{units} and the direction is 35.23^{\circ}

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