Is the Force balanced or unbalanced?
1 answer:
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Answer:
1.7323
Explanation:
To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.
From the data given we have to:
Where n means the index of refraction.
We need to calculate the index of refraction of the liquid, then applying Snell's law we have:
Replacing the values we have:
Therefore the refractive index for the liquid is 1.7323
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)
q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
Answer:
Explanation:
Let the linear charge density of the charged wire is given as
here we can use Gauss law to find the electric field at a distance r from wire
so here we will assume a Gaussian surface of cylinder shape around the wire
so we have
here we have
so we have
<h2>Hello!</h2>
The answer is: 19.59 m
<h2>Why?</h2>Since there is no information about the launch type, we can assume that the ball is thrown vertically upward.
When the ball reaches the maximum height, just at that moment, the velocity turns to 0, and after that moment, the ball starts falling, so:
We will use the following formula:
Where:
Vf= Final velocity = 0
Vi= Initial velocity =
g = Gravity Acceleration =
s = Traveled distance
Have a nice day!