The students on the right are applying a force of 100 N to the right. This is shown by the red arrow. The students on the left a
re applying a force of 100 N to the left. This is shown by the blue arrow. A student on the blue team loses balance and lets go of the rope. What is the most likely effect of this change?
A. The rope will move to the right, in the direction of the net force.
B. The rope will move to the left, in the direction opposite to the net force.
C. The rope will move to the left, in the direction of the net force.
D. The rope will move to the right, in the direction opposite to the net force.
A. The rope will move to the right, in the direction of the net force.
B. The rope will move to the left, in the direction opposite to the net force.
C. The rope will move to the left, in the direction of the net force.
D. The rope will move to the right, in the direction opposite to the net force.
2 answers:
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Answer:
D₂= 167,21 cm : Magnitude of the second displacement
β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement
Explanation:
We find the x-y components for the given vectors:
i: unit vector in x direction
j:unit vector in y direction
D₁: Displacement Vector 1
D₂: Displacement Vector 2
R= resulta displacement vector
D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j
D₂= -D₂(i)-D₂(j)
R= 131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j
We propose the vector equation for sum of vectors:
D₁+ D₂= R
-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j
-51.99i+D₂x(i)=103.23i
D₂x=103.23+51.99=155.22 cm
+142.83j-D₂y(j) =+80.65j
D₂y=142.83-80.65=62.18 cm
Magnitude and direction of the second displacement
D₂= 167.21 cm
Direction of the second displacement
β= 21.8°
D₂= 167,21 cm : Magnitude of the second displacement
β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement
