a 2.0 kg block on an incline at a 60.0 degree angle is held in equilibrium by a horizontal force, what is the magnitude of this
2 answers:
Answer:
Horizontal force is 16.97 N.
Explanation:
It is given that,
Mass of the block, m = 2 kg
It is on an incline at a 60 degree angle is held in equilibrium by a horizontal force. We need to find the magnitude of horizontal force. The force acting on the block is its weight mg.
The horizontal component of force is,
The vertical component of force is,
So, the horizontal component of force is,
So, the horizontal component of force is 16.97 N.
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The electric field produced by a single-point charge is given by
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.
1) The first charge is , and it is located at x=0, so its distance from the point x=0.200 m is
Therefore, the electric field is
And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.
2) The second charge is and it is located at x=0.800 m, so its distance from the point is
Therefore, the electric field is
And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.
3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have
and the sign tells us that the field is directed toward negative x-direction.
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
