Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis
![F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}](https://tex.z-dn.net/?f=F_N%3Dk_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D)
Explanation:
Let the force on +Q charge y-axis due to +2Q charge be 
and force on +Q charge y axis due to -Q charge on x-axis be 
.
Distance between the +2Q charge and +Q charge = d units
Distance between the -Q charge and +Q charge = 
units
= Coulomb constant
Net force on +Q charge on y-axis is:
![|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|](https://tex.z-dn.net/?f=%7CF_N%7C%3D%7Ck_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D%7C)
The net froce on the +Q charge on y-axis is
Answer:
Can you give a clearer question
Explanation:
pls give me points i have questions
Answer:
y₁ = 37.2 m, y₂ = 22,6 m
Explanation:
For this exercise we can use the kinematic equations
For the globe, with index 1
y₁ = y₀ + v₁ t
For the shot with index 2
y₂ = 0 + v₂ t - ½ g t²
At the point where the position of the two bodies meet is the same
y₁ = y₂
y₀ + v₁ t = v₂ t - ½ g t²
14 + 8.40t = 27.0 t - ½ 9.8 t²
4.9 t² - 18.6 t + 14 = 0
t² - 3,796 t + 2,857 = 0
Let's look for time by solving the second degree equation
t = [3,796 ±√(3,796 2 - 4 2,857)] / 2
t = [3,796 ± 1,727] / 2
t₁ = 2.7615 s
t₂ = 1.03 s
Now we can calculate the distance for each time
y₁ = v₂ t₁ - ½ g t₁²
y₁ = 27 2.7615 - ½ 9.8 2.7615²
y₁ = 37.2 m
y₂ = v₂ t₂ - ½ g t₂²
y₂ = 27 1.03 - ½ 9.8 1.03²
y₂ = 22,612 m
