Respuesta :
Answer:
The magnitude of the resultant decreases from A+B to A-B
Explanation:
The magnitude of the resultant of two vectors is given by
[tex]R=\sqrt{A^2 +B^2 +2AB cos \theta}[/tex]
where
A is the magnitude of the first vector
B is the magnitude of the second vector
[tex]\theta[/tex] is the angle between the directions of the two vectors
In the formula, A and B are constant, so the behaviour depends only on the function [tex]cos \theta[/tex]. The value of [tex]cos \theta[/tex] are:
- 1 (maximum) when the angle is 0, so the magnitude of the resultant in this case is
[tex]R=\sqrt{A^2 +B^2+2AB}=\sqrt{(A+B)^2}=A+B[/tex]
- then it decreases, until it becomes 0 when the angle is 90 degrees, where the magnitude of the resultant is
[tex]R=\sqrt{A^2 +B^2+0}=\sqrt{A^2+B^2}[/tex]
- then it becomes negative, and continues to decrease, until it reaches a value of -1 when the angle is 180 degrees, and the magnitude of the resultant is
[tex]R=\sqrt{A^2 +B^2-2AB}=\sqrt{(A-B)^2}=A-B[/tex]
The resultant of the two vectors when the angle between them is increased from 0 to 180 is 12 N.
The given parameters;
- first force, a = 5 N
- second force, b = 7 N
The resultant of the two vectors will be determined using parallelogram rule as shown below;
[tex]R^2 = a^2 + b^2 - 2ab\ cos (\theta)[/tex]
where;
- R is the resultant vector
R² = (5)² + (7)² - (2 x 5 x 7) cos(180)
R² = 74 - (-70)
R² = 74 + 70
R² = 144
R = √144
R = 12 N
Thus, the resultant of the two vectors when the angle between them is increased from 0 to 180 is 12 N.
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