Respuesta :
Answer:
You're supposed to measure the distance from X to the end of vector 5 using the appropriate scale, and measure the angle (counterclockwise from X) using a protractor.
Mathematically:
x = [83cos90+55cos141+69cos229+41cos281+61co... cm = -27 cm
y = [83sin90+55sin141+69sin229+41sin281+61si... cm = 55 cm
so d = √(x² + y²) = 61 cm
and Θ = arctan(55/-27) = -64º +180º (to get into QII) = 116º
Explanation:
The magnitude of the resultant vectors is 64.69 cm
The direction of the vector of the vector is 332.4 ⁰ north west
The given parameters;
[tex]displacement \ at \ 90^0 = 83 \ cm\\\\displacement \ at \ 147^0 = 59 \ cm\\\\displacement \ at \ 221^0 = 69 \ cm\\\\displacement \ at \ 283^0 = 45 \ cm\\\\displacement \ at \ 27^0 = 69 \ cm[/tex]
The resultant displacement is calculated by resolving each vector into x and y components.
The y-component of the vectors assuming all vectors are measured counter-clockwise from x-axis;
[tex]R_y = (83\times sin(90)) + (59\times sin(147)) + (69\times sin(221)) + (45\times sin(283))\\ \ \ \ . \ \ \ \ \ \ + (69\times sin(27)) \\\\R_y = (83) + (32.13) + (-45.27) + (-43.85) + (31.33)\\\\R_y = (57.34) j[/tex]
The x-component of the vectors
[tex]R_x = (83\times cos(90)) + (59\times cos(147)) + (69\times cos(221)) + (45\times cos(283))\\ \ \ \ . \ \ \ \ \ \ + (69\times cos(27)) \\\\R_x = (0) + (-49.48) + (-52.07) + (10.12) + (61.48)\\\\R_x = (-29.95) i[/tex]
The magnitude of the resultant vectors is calculated as;
[tex]|R| = \sqrt{R_y^2 + R_x^2} \\\\|R| = \sqrt{(57.34)^2 + (-29.95)^2} \\\\|R| = 64.69 \ cm[/tex]
The direction of the vector;
[tex]\theta = tan^{-1} (\frac{R_y}{R_x} )\\\\\theta = tan^{-1} (\frac{57.34}{-29.95} )\\\\\theta = tan^{-1} (-1.915) = -62.4^0 = 332.4 ^0 \ north \ west[/tex]
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