Answer:
![F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]](https://tex.z-dn.net/?f=F_T%3D6k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bi%7D%2B10k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bj%7D%3D2k%5Cfrac%7BQ%5E2%7D%7BL%7D%5B3%5Chat%7Bi%7D%2B5%5Chat%7Bj%7D%5D)


Explanation:
I attached an image below with the scheme of the system:
The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:
![F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]](https://tex.z-dn.net/?f=F_T%3DF_Q%2BF_%7B3Q%7D%2BF_%7B4Q%7D%5C%5C%5C%5CF_T%3Dk%5Cfrac%7B%28Q%29%282Q%29%7D%7BR_1%7D%5Chat%7Bi%7D%2Bk%5Cfrac%7B%283Q%29%282Q%29%7D%7BR_2%7D%5Chat%7Bj%7D%2Bk%5Cfrac%7B%284Q%29%282Q%29%7D%7BR_3%7D%5Bcos%5Ctheta%20%5Chat%7Bi%7D%2Bsin%5Ctheta%20%5Chat%7Bj%7D%5D)
the distances R1, R2 and R3, for a square arrangement is:
R1 = L
R2 = L
R3 = (√2)L
θ = 45°
![F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]](https://tex.z-dn.net/?f=F_T%3Dk%5Cfrac%7B2Q%5E2%7D%7BL%7D%5Chat%7Bi%7D%2Bk%5Cfrac%7B6Q%5E2%7D%7BL%7D%5Chat%7Bj%7D%2Bk%5Cfrac%7B8Q%5E2%7D%7B%5Csqrt%7B2%7DL%7D%5Bcos%2845%5C%C2%B0%29%5Chat%7Bi%7D%2Bsin%2845%5C%C2%B0%29%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_T%3Dk%5Cfrac%7B2Q%5E2%7D%7BL%7D%5Chat%7Bi%7D%2Bk%5Cfrac%7B6Q%5E2%7D%7BL%7D%5Chat%7Bj%7D%2Bk%5Cfrac%7B8Q%5E2%7D%7B%5Csqrt%7B2%7DL%7D%5B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Chat%7Bi%7D%2B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_T%3D6k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bi%7D%2B10k%5Cfrac%7BQ%5E2%7D%7BL%7D%5Chat%7Bj%7D%3D2k%5Cfrac%7BQ%5E2%7D%7BL%7D%5B3%5Chat%7Bi%7D%2B5%5Chat%7Bj%7D%5D)
and the magnitude is:

the direction is:

The initial velocity of the ball is 1.01 m/s
Explanation:
The motion of the ball rolling off the desk is a projectile motion, which consists of two independent motions:
- A uniform horizontal motion with constant horizontal velocity
- A vertical accelerated motion with constant acceleration (
, acceleration due to gravity)
We start by analyzing the vertical motion: we can find the time of flight of the ball by using the following suvat equation

where
s = 1.20 m is the vertical displacement (the height of the desk)
u = 0 is the initial vertical velocity

t is the time of flight
Solving for t,

Now we analyze the horizontal motion. We know that the ball covers a horizontal distance of
d = 0.50 m
in a time
t = 0.495 s
Therefore, since the horizontal velocity is constant, we can calculate it as

So, the ball rolls off the table at 1.01 m/s.
Learn more about projectile motion:
brainly.com/question/8751410
#LearnwithBrainly
Answer:
1.6 kg
Step-by-step Solution:
Since Force = mass × acceleration we have:
F = 8N
a= 5 m/s^2
m = ?
By plugging the values above into F=ma we obtain:

Therefore, the Chromebook has a mass of 1.6 kilograms.
Since direction is part of velocity, and the object is moving in a circle, its velocity is constantly changing.
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