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:
Answer:
The magnification produced by a plane mirror is +1
means then the size of the image is equal to the size of the object. If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object.
Answer:
PE = (|accepted value – experimental value| \ accepted value) x 100%
Explanation:
The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
Which means that the frequency is
and the angular frequency is
In a spring-mass system, the maximum velocity of the object is given by
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
Answer:
motion energy to sound energy
Explanation:
you move your hands together, then clap then, which makes a sound. Hope that this helps you and have a great day :)
