Se aplican dos fuerzas concurrentes a un objeto de 4N a la derecha y 5N a la izquierda. ¿Hacia donde se movió y con cuanta fuerz
1 answer:
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Answer:
- This means that the integral of the square modulus over the space is dimensionless.
Explanation:
We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral
must be dimensionless, as represents a probability.
As the differentials has units of length
for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:
taking the square root this gives us :
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
