1. In Physics, we distinguish between wave motion and particle motion Wave motion
1 answer:
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Answer:
d’= (0.561 i ^ - 0.634 j ^) m , d’= 0.847 m , 48.5 south east
Explanation:
This is a displacement exercise, one of the easiest methods to solve it is to decompose the displacements in a coordinate system. Let's start with beetle 1
Let's use trigonometry to break down your second displacement
d₂ = 0.89 m θ = 32 north east
sin θ = / d₂
d_{2y} = d2 sin 32
d_{2y} = 0.89 sin 32
d_{2y} = 0.472 m
cos 32 = d₂ₓ / d₂
d₂ₓ = d₂ cos 32
d₂ₓ = 0.89 cos 32
d₂ₓ = 0.755 m
We found the total displacement of the beetle 1
X axis
d₁ = 0.58 i ^
Dₓ = d₁ + d₂ₓ
Dₓ = 0.58 + 0.755
Dₓ = 1,335 m
Axis y
D_{y} = d_{2y}
D_{y} = 0.472 m
Now let's analyze the second beetle
d₃ = 1.37 m θ = 35 north east
Sin (90-35) = d_{3y} / d₃
d_{3y} = d₃ sin 55
d_{3y} = 1.35 sin 55
d_{3y} = 1,106 m
cos 55 = d₃ₓ / d₃
d₃ₓ = d₃ cos 55
d₃ₓ = 1.35 cos 55
d₃ₓ = 0.774 m
They ask us what the second displacement should be to have the same location as the beetle 1
Dₓ = d₃ₓ + dx’
D_{y} = d_{3y} + dy’
dx’= Dₓ - d₃ₓ
dx’= 1.335 - 0.774
dx’= 0.561 m
dy’= D_{y} - d_{3y}
dy’= 0.472 - 1,106
dy’= -0.634 m
We can give the result in two ways
d’= (0.561 i ^ - 0.634 j ^) m
Or in the form of module and address
d’= √ (dx’² + dy’²)
d’= √ (0.561² + 0.634²)
d’= 0.847 m
tan θ = dy’/ dx’
θ = tan⁻¹ dy ’/ dx’
θ = tan⁻¹ (-0.634 / 0.561)
θ = -48.5 º
This is 48.5 south east