Answer:
X₃₁ = 0.58 m and X₃₂ = -1.38 m
Explanation:
For this exercise we use Newton's second law where the force is the Coulomb force
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
Since all charges are of the same sign, forces are repulsive
F₁₃ = k q₁ q₃ / r₁₃²
F₂₃ = k q₂ q₃ / r₂₃²
Let's find the distances
r₁₃ = x₃- 0
r₂₃ = 2 –x₃
We substitute
k q q / x₃² = k 4q q / (2-x₃)²
q² (2 - x₃)² = 4 q² x₃²
4- 4x₃ + x₃² = 4 x₃²
5x₃² + 4 x₃ - 4 = 0
We solve the quadratic equation
x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5
x₃ = [-4 ± 9.80] 10
X₃₁ = 0.58 m
X₃₂ = -1.38 m
For this two distance it is given that the two forces are equal
Answer:

Explanation:
The magnitude of the net force exerted on q is known, we have the values and positions for
and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by
on q. Then we can know the magnitude of the force exerted by
about q, finally this will allow us to know the magnitude of 
exerts a force on q in +y direction, and
exerts a force on q in -y direction.

The net force on q is:

Rewriting for
:

Answer:
B. Solids, liquids, and gases.
Explanation:
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Answer:
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Answer:
temperature on left side is 1.48 times the temperature on right
Explanation:
GIVEN DATA:

T1 = 525 K
T2 = 275 K
We know that


n and v remain same at both side. so we have

..............1
let final pressure is P and temp 

..................2
similarly
.............3
divide 2 equation by 3rd equation
![\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B11%7D%5E%7B-2%2F3%7D%20%5Cfrac%7B21%7D%7B11%7D%5E%7B5%2F3%7D%20%3D%20%5B%5Cfrac%7BT_1%20%7Bf%7D%7D%7BT_2%20%7Bf%7D%7D%5D%5E%7B5%2F3%7D)

thus, temperature on left side is 1.48 times the temperature on right