Respuesta :
Answer:
[tex]|B|=27.00425726m[/tex]
[tex]\alpha =210.3781372[/tex]°
Explanation:
Let's use the component method of vector addition:
[tex]A_x=13.6cos(40.2)=10.38762599\\A_y=13.6sin(40.2)=8.778224553\\Cx=13.8cos(20.7+180)=-12.90912763\\Cy=13.8sin(20.7+180)=-4.877952844[/tex]
Now, we know:
[tex]C_x=A_x+B_x\\\\C_y=A_y+b_y[/tex]
So:
[tex]B_x=C_x-A_x=-23.29675362\\B_y=C_y-A_y=-13.6561774[/tex]
Now lets calculate the magnitude of the vector B:
[tex]|B|=\sqrt{(B_x)^{2} +(B_y)^{2} }=27.00425726m[/tex]
Finally its angle is given by:
[tex]\alpha =(arctan(\frac{B_y}{B_x}))+180=30.37813438+180=210.3781344[/tex]°
Keep in mind that I added 180 to the angles of C and B to find the real angles measured from the + x axis counter-clock wise.
Vector is quantity. The magnitude of vector B is 4.644 m while the angle from the positive x-direction is 302.88°.
What is Vector?
A Vector is a quantity in physics that has both magnitude and direction.
We know that in order to add to vector we need to divide the vector into two parts, a sine(Vertical) and a cosine(Horizontal), therefore,
The vertical addition of the vectors A and B can be written as,
[tex]\vec A_y +\vec B_y = \vec C_y[/tex]
[tex]\vec A(Sin\ \theta_A) +\vec B(Sin\ \theta_B) = \vec C(Sin\ \theta_C)[/tex]
[tex]13.6(Sin\ 40.2^o) +\vec B(Sin\ \theta_B) = 13.8(Sin\ 20.7^o)\\\\\vec B(Sin\ \theta_B ) =-3.9[/tex]
The Horizontal addition of the vectors A and B can be written as,
[tex]\vec A_x +\vec B_x = \vec C_x[/tex]
[tex]\vec A(Cos\ \theta_A) +\vec B(Cos\ \theta_B) = \vec C(Cos\ \theta_C)[/tex]
[tex]13.6(Cos\ 40.2^o) +\vec B(Cos\ \theta_B) = 13.8(Cos\ 20.7^o)\\\\\vec B(Cos\ \theta_B ) =2.5215[/tex]
As the value of Sin is negative and the value of Cos is positive, therefore, Vector B will lie in the fourth quadrant.
The angle of Vector B,
[tex]\dfrac{\vec B\ Sin\ \theta_B}{\vec B\ Cos\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\\dfrac{\ Sin\ \theta_B}{\ Cos\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\Tan\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\\theta_B = Tan^{-1}\ \dfrac{-3.9}{2.5215}\\\\\theta_B = -57.12^o[/tex]
Thus, the angle of vector B is 57.12° clockwise from the -x direction.
In order to make the angle positive, we can deduct the value from 360°,
[tex]\theta_B = -57.12^o\\\\\theta_B = 360^o -57.12^o\\\\\theta_B = 302.88^o[/tex]
The magnitude of Vector B,
We know the value of the Perpendicular component of vector B,
[tex]\vec B(Sin\ \theta_B ) =-3.9\\\\\vec B(Sin\ -57.12^o ) =-3.9\\\\\vec B= 4.644\rm\ m[/tex]
Hence, the magnitude of vector B is 4.644 m while the angle from the positive x-direction is 302.88°.
Learn more about Vector:
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