Answer:
447 K
Explanation:
25 C = 25 + 273 = 298 K
Assuming ideal gas, we can apply the ideal gas law
![\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%20%3D%20%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
![T_2 = T_1\frac{P_2}{P_1}\frac{V_2}{V_1}](https://tex.z-dn.net/?f=T_2%20%3D%20T_1%5Cfrac%7BP_2%7D%7BP_1%7D%5Cfrac%7BV_2%7D%7BV_1%7D)
Since pressure is tripled, then
. Volume is halved, then ![V_2 / V_1 = 0.5](https://tex.z-dn.net/?f=V_2%20%2F%20V_1%20%3D%200.5)
![T_2 = 298*3*0.5 = 447 K](https://tex.z-dn.net/?f=T_2%20%3D%20298%2A3%2A0.5%20%3D%20447%20K)
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
Sedimentary rock I believe
Answer: B
Explanation: This can be easily done by inputting 5280 * 2.2.