The height of a transverse waves is known as its?
1 answer:
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Answer:
A+B; 5√5 units, 341.57°
A-B; 5√5 units, 198.43°
B-A; 5√5 units, 18.43°
Explanation:
Given A = 5 units
By vector notation and the axis of A, it is represented as -5j
B = 3 × 5 = 15 units
Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B
(a) A + B = -5j + 15i
A+B = 15i -5j
|A+B| = √(15)²+(5)²
= 5√5 units
∆ = arctan(5/15) = 18.43°
The angle ∆ is generally used in the diagrams
∆= 18.43°
The direction of A+B is 341.57° based in the condition given (see attachment for diagrams
(b) A - B = -5j -15i
A-B = -15i -5j
|A-B|= √(15)²+(-5)²
|A-B| = √125
|A-B| = 5√5 units
The direction is 180+18.43°= 198.43°
See attachment for diagrams
(c) B-A = 15i -( -5j) = 15i + 5j
|B-A| = 5√5 units
The direction is 18.43°
See attachment for diagram
