Answer:
29.4 uN
Explanation:
The electric force between two charges can be calculated using Coulomb's Law. According to this law the force between two point charges is given as:
where k is a proportionality constant known as the Coulomb's law constant. Its value is 
Nm²/C²
r = distance between charges = 70 cm = 0.7 m
q1 = q2 = 4nC = 
C
The negative sign indicates that the charges are negative. In the formula we will only use the magnitude of the charges.
Using these values in the formula, we get:
Therefore, the magnitude of repulsive force between the given charges will be 29.4 uN
Explanation:
The magnitude of the electric field between the plates is given by
E = -ΔV/d
minus sign indicates Potential decreases in the direction of electric field
where
ΔV is the potential difference between the plates
D is the distance between the plates.
The work done when carrying an electrical charge on an equipotential surface between one position to the other is zero W= q(V-V)=0 The electric field lines of force are always perpendicular to an equipotential surface. That conductor in an equipotential surface as direction E is at right angles to an eauipotential surface The intensity of the electric field along an equipotential surface is always zero. Equipotential surfaces never collide with each other as this would mean that at that point, there are two alternative values that are not true.
Answer:
Explanation:
A function f(x) is a Probability Density Function if it satisfies the following conditions:
Given the function:
(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in 
(2)
![\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20p%28x%29%3D%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20%5Cdfrac%7B1%7D%7Br%7De%5E%7B-x%2Fr%7D%5C%5C%3D%5Cdfrac%7B1%7D%7Br%7D%20%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20e%5E%7B-x%2Fr%7D%5C%5C%3D-%5Cdfrac%7Br%7D%7Br%7D%5Cleft%5Be%5E%7B-x%2Fr%7D%5Cright%5D_%7B0%7D%5E%7B%5Cinfty%7D%5C%5C%3D-%5Cleft%5Be%5E%7B-%5Cinfty%2Fr%7D-e%5E%7B-0%2Fr%7D%5Cright%5D%5C%5C%3D-e%5E%7B-%5Cinfty%7D%2Be%5E%7B-0%7D%5C%5CSInce%20%5C%3A%20e%5E%7B-%5Cinfty%7D%20%5Crightarrow%200%5C%5Ce%5E%7B-0%7D%3D1%5C%5C%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20p%28x%29%3D1)
The function p(x) satisfies the conditions for a probability density function.
Yes it can because it had lots of force
Below are the choices that can be found from other source:
a. frequency
b. the Doppler effect
c. the resonance effect
<span>d. intensity
The answer is B.
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