<span>
A convection current is a flow of a fluid or air due to heating and
cooling of the fluid or air. And for that matter the IE: Hot air rises. Cold air falls.</span>
To solve this problem we will apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation for which we can find the speed of this
![V = \frac{kq}{r}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bkq%7D%7Br%7D)
Here,
k = Coulomb's constant
q = Charge
r = Distance to the center point between the charge
From each object the potential will be
![V_1 = \frac{kq_1}{r_1}+\frac{kq_2}{r_2}](https://tex.z-dn.net/?f=V_1%20%3D%20%5Cfrac%7Bkq_1%7D%7Br_1%7D%2B%5Cfrac%7Bkq_2%7D%7Br_2%7D)
Replacing the values we have that
![V_1 = \frac{(9*10^9)(3.05*10^{-9})}{0.41/2}+\frac{(9*10^9)(1.85*10^{-9})}{0.41/2}](https://tex.z-dn.net/?f=V_1%20%3D%20%20%5Cfrac%7B%289%2A10%5E9%29%283.05%2A10%5E%7B-9%7D%29%7D%7B0.41%2F2%7D%2B%5Cfrac%7B%289%2A10%5E9%29%281.85%2A10%5E%7B-9%7D%29%7D%7B0.41%2F2%7D)
![V_1 = 215.12V](https://tex.z-dn.net/?f=V_1%20%3D%20215.12V)
Now the potential two is when there is a difference at the distance of 0.1 from the second charge and the first charge is 0.1 from the other charge, then,
![V_1 = \frac{(9*10^9)(3.05*10^{-9})}{0.1}+\frac{(9*10^9)(1.85*10^{-9})}{0.41-0.1}](https://tex.z-dn.net/?f=V_1%20%3D%20%20%5Cfrac%7B%289%2A10%5E9%29%283.05%2A10%5E%7B-9%7D%29%7D%7B0.1%7D%2B%5Cfrac%7B%289%2A10%5E9%29%281.85%2A10%5E%7B-9%7D%29%7D%7B0.41-0.1%7D)
![V_2 = 328.2V](https://tex.z-dn.net/?f=V_2%20%3D%20328.2V)
Applying the energy conservation equations we will have that the kinetic energy is equal to the electric energy, that is to say
![\frac{1}{2} mv^2 = q(V_2-V_1)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%3D%20q%28V_2-V_1%29)
Here
m = mass
v = Velocity
q = Charge
V = Voltage
Rearranging to find the velocity
![v = \sqrt{ \frac{2q(V_2-V_1)}{m}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%20%5Cfrac%7B2q%28V_2-V_1%29%7D%7Bm%7D%7D)
Replacing,
![v = \sqrt{ \frac{-2(1.6*10^{-19})(328.2-215.12)}{9.11*10^{-3}}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%20%5Cfrac%7B-2%281.6%2A10%5E%7B-19%7D%29%28328.2-215.12%29%7D%7B9.11%2A10%5E%7B-3%7D%7D%7D)
![v = 6.3*10^6m/s](https://tex.z-dn.net/?f=v%20%3D%206.3%2A10%5E6m%2Fs)
Therefore the speed final velocity of the electron when it is 10.0 cm from charge 1 is ![6.3*10^6m/s](https://tex.z-dn.net/?f=6.3%2A10%5E6m%2Fs)
Answer:
On a velocity-time graph… slope is acceleration. the "y" intercept is the initial velocity. when two curves coincide, the two objects have the same velocity at that time.
The sun light that received by the water in the ocean will increase the average temperature of the water and make it warmer.
With the help of wind and current, the warm water will spread out to another region, and increasing the average temperature in that region and affecting its overall climate.
Example The temperature rise in Valdivia<span>, Chile and in Beijing, China after receiving warm water from arctic.</span>
To solve this problem we will apply the linear motion kinematic equations. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,
![h = v_0 t -\frac{1}{2} gt^2](https://tex.z-dn.net/?f=h%20%3D%20v_0%20t%20-%5Cfrac%7B1%7D%7B2%7D%20gt%5E2)
![-18 = 15*t + \frac{1}{2} 9.8*t^2](https://tex.z-dn.net/?f=-18%20%3D%2015%2At%20%2B%20%5Cfrac%7B1%7D%7B2%7D%209.8%2At%5E2)
![t = 3.98s](https://tex.z-dn.net/?f=t%20%3D%203.98s)
Then the total distance traveled would be
![h = h_0 +v_0t](https://tex.z-dn.net/?f=h%20%3D%20h_0%20%2Bv_0t)
![h = 18+15*3.98](https://tex.z-dn.net/?f=h%20%3D%2018%2B15%2A3.98)
![h = 77.7m](https://tex.z-dn.net/?f=h%20%3D%2077.7m)
Therefore the railing will be at a height of 77.7m when it has touched the ground