Part A
Free fall motion
h = 3.1 m
Equation: Vf = √(2gh) = √(2*9.8 m/s^2 * 3.1 m) = 7.8 m/s
That is the only part in the question.
Answer:
0.247 μC
Explanation:
As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:
![F_y: T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N](https://tex.z-dn.net/?f=F_y%3A%20%20T_y%20-%20W%20%3D%200%5C%5CT_y%20%3D%20m%2Ag%20%3D%200.002%20kg%20%2A9.81m%2Fs%5E2%20%3D%200.01962%20N)
![T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N](https://tex.z-dn.net/?f=T_y%20%3D%20T_%2Acos%2850%29%5C%5CT%20%3D%20%5Cfrac%7BT_y%7D%7Bcos%2850%29%7D%20%3D%200.0305%20N)
![T_x = T*sin(50) = 0.0234 N](https://tex.z-dn.net/?f=T_x%20%3D%20T%2Asin%2850%29%20%3D%200.0234%20N)
The electric force is given by the expression:
![F = k*\frac{q_1*q_2}{r^2}](https://tex.z-dn.net/?f=F%20%3D%20k%2A%5Cfrac%7Bq_1%2Aq_2%7D%7Br%5E2%7D)
In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):
![r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m](https://tex.z-dn.net/?f=r%20%3D%202%2AL%2Asin%2850%29%20%3D%202%20%2A%200.1m%20%2A%20sin%2850%29%200.1532%20m)
And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.
![F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}](https://tex.z-dn.net/?f=F_x%20%3D%20T_x%20-%20F_e%20%3D%200%5C%5CT_x%20%3D%20F_e%20%3D%20k%2A%5Cfrac%7Bq%5E2%7D%7Br%5E2%7D)
![q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C](https://tex.z-dn.net/?f=q%20%3D%20%5Csqrt%7BT_x%20%2A%5Cfrac%7Br%5E2%7D%7Bk%7D%7D%20%3D%20%5Csqrt%7B0.0234%20N%20%2A%20%5Cfrac%7B%280.1532m%29%5E2%7D%7B9%2A10%5E9%20N%2Am%5E2%2FC%5E2%7D%20%7D%20%3D%202.4704%20%2A%2010%5E-7%20C)
O 0.247 μC
Answer: The pressure that one experiences on the Mount Everest will be different from the one, in a classroom. It is because pressure and height are inversely proportional to each other. This means that as we move up, the height keeps on increasing but the pressure will keep on decreasing. This is the case that will be observed when one stands on the Mount Everest as the pressure is comparatively much lower there.
It is because as we move up, the amount of air molecules keeps on decreasing but all of the air molecules are concentrated on the lower part of the atmosphere or on the earth's surface.
Thus a person in a low altitude inside a classroom will experience high pressure and a person standing on the Mount Everest will experience low pressure.