Hi there!
Recall the equation for electric potential of a point charge:
![V = \frac{kQ}{r}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7BkQ%7D%7Br%7D)
V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)
Q = Charge (C)
r = distance (m)
We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.
Upper right charge's potential:
![V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B%288.99%2A10%5E9%29%28-5%20%2A%2010%5E%7B-9%7D%29%7D%7B0.04%7D%20%3D%20-1123.75%20V)
Lower left charge's potential:
![V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B%288.99%2A10%5E9%29%285%2A10%5E%7B-9%7D%29%7D%7B0.02%7D%20%3D%202247.5%20V)
Add the two, and subtract from the total EP at the point:
![3160 + 1123.75 - 2247.5 = 2036.25](https://tex.z-dn.net/?f=3160%20%2B%201123.75%20-%202247.5%20%3D%202036.25)
The remaining charge must have a potential of 2036.25 V, so:
![2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}](https://tex.z-dn.net/?f=2036.25%20%3D%20%5Cfrac%7B%288.99%2A10%5E9%29%28Q%29%7D%7B%5Csqrt%7B0.02%5E2%20%2B%200.04%5E2%7D%7D%5C%5C%5C%5C2036.25%20%3D%20%5Cfrac%7B%288.99%2A10%5E9%29Q%7D%7B0.0447%7D%20%5C%5C%5C%5CQ%20%3D%200.000000010127%20%3D%20%5Cboxed%7B10.13nC%7D)
I don't really know the answer but maybe north pole and south pole?
Work = force * distance
and newton*meters = Joule
In this case, work = 250N*50m = 12500 J
So the answer is D) 12,500 J
Answer:
aₓ = 0
, ay = -6.8125 m / s²
Explanation:
This is an exercise that we can solve with kinematics equations.
Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.
x axis
vₓ = v₀ₓ = 1.10 m / s
aₓ = 0
y axis
initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration
= v_{oy} -ay t
ay = (v_{oy} -v_{y}) / t
ay = (0 -10.9) / 1.6
ay = -6.8125 m / s²
the sign indicates that the acceleration goes in the negative direction of the y axis