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.
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The answer is a. generator!
Answer:
Magnitude = 3.64 × 
စ = 43.9°
Explanation:
given data
ship to travel = 1.7 × kilometers
turn = 70°
travel an additional = 2.7 × kilometers
solution
we will consider here
Px = 1.7 ×
Py = 0
Qx =2.7 × cos(70)
Qy= 2.7 × sin(70)
so that
Hx = Px + Qx ............1
Hx = 2.62 ×
and
Hy = Py + Qy ..........2
Hy = 2.53 ×
so Magnitude = 
Magnitude = 3.64 ×
so direction will be
tan စ = Hy ÷ Hx ......................3
tan စ = 
tan စ = 0.9656
စ = 43.9°
The insect's velocity is 0.2 meter/second along east.
<h3>What is velocity?</h3>
The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.
Given parameters:
An insect crawls 4 m east along a hiking path.
Time taken by it: t = 20 seconds.
We have to fine: the insect's velocity, v =?
So, the magnitude of velocity: v= 4/20 meter/second = 0.2 meter/second.
And the direction velocity is along east.
Hence, the insect's velocity is 0.2 meter/second along east.
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The correct answer is
<span>Density
In fact, the density of an object is the ratio between its mass m and its volume V:
</span>
<span>if the size of the object changes, its volume V changes as well, therefore the density d must change.</span>
