Answer:resultant vector R = (0, 3)
Explanation: vector A = (3, 0)
vector B =(-3, 3)
Vectors are added such that those in same directions are added together. The resultant vector R is the given by R = (3-3, 0+3)
= (0, 3)
Answer:
Depending on which hemisphere it is, like western to eastern, It would most likely get stuck at the center. You would also have to put more things into thought like acceleration, velocity, and speed.
BUT since the question asked "would it pop out the other side?", I'm assuming it's talking about northern to southern hemisphere. so in that case it would pop out the other side since gravity makes things go downwards.
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:
Step 1: Identify the variables. ...Step 2: Determine the variable range. ...Step 3: Determine the scale of the graph. ...Step 4: Number and label each axis and title the graph.Step 5: Determine the data points and plot on the graph. ...Step 6: Draw the graph.
Acceleration is the rate at which an object picks up speed. deceleration is the rate at which an object loses speed.
