The impact speed will be
v^2 = 2*9.8*1.3
v^2 = 25.48
v= 5.04 m/s
The electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.
<h3>
Electric potential energy</h3>
When work is done on a positive test charge to move it from one location to another, potential energy increases and electric potential increases.
The electric potential energy between the charges when the second charge is at point b is calculated as follows;
ΔU = -w
Ui - Uf = w
Uf = Ui - w
where;
Uf is the final potential energy
Ui is the initial potential energy
w is the work done by the force
Uf = 5.4 x 10⁻⁸ J - (-1.9 x 10⁻⁸J)
Uf = 5.4 x 10⁻⁸ J + 1.9 x 10⁻⁸ J
Uf = 7.3 x 10⁻⁸ J
Thus, the electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.
Learn more about electric potential energy here: brainly.com/question/14306881
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I just figured this out now.
First you would use the formula
Ephoton= hc/λ and substitute in the value's of plank's constant, the speed of light in a vaccum and the wavelength which will give you the energy in joules. Then you go to the reference table and solve for the energy used between the different levels for Mercury making sure to convert electron volts to jules. In the end the correct answer should be energy level D.
Answer:
Therefore,
The speed of the wave on the longer wire is 95 m/s.
Explanation:
Given:
For Short wire, speed is
Let length of Short and Longer wire be 
such that
To Find:
Speed on the longer wire
Solution:
The speed of a pulse or wave on a string under tension can be found with the equation,
Where,
= Tension on the wire
L = Length of Sting
m = mass of String
So here we have,
= same
Therefore,
......equation ( 1 )
And
.......equation ( 2 )
Dividing equation 1 by equation 2 and on Solving we get
Therefore,
Therefore,
The speed of the wave on the longer wire is 95 m/s.
