Answer:

Explanation:
Given data
Electric potential at point a is Ua=5.4×10⁻⁸J
q₂ moves to point b where a negative work done on it
Required
Electric potential energy Ub
Solution
When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

Now substitute the given values
So

Answer:
C. rollback
Explanation:
The retina of the eye is located inside the eye making it impossible for the retina to roll back. Only the eye itself can rollback.
Answer:
The tension is 
Explanation:
The free body diagram of the question is shown on the first uploaded image From the question we are told that
The distance between the two poles is 
The mass tied between the two cloth line is 
The distance it sags is 
The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline
Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium
And this can be mathematically represented as

To obtain
we apply SOHCAHTOH Rule
So 
![\theta = tan^{-1} [\frac{opp}{adj} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7Bopp%7D%7Badj%7D%20%5D)
![= tan^{-1} [\frac{1}{7}]](https://tex.z-dn.net/?f=%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7B7%7D%5D)






Answer:
Displacement: 6.71 m, Direction: 63.4 degrees north of east
Explanation:
In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).
Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).
And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)
And the displacement will be defined by the folliwing vector operation:

We know that the magnitude of the displacement vector is defined by the phytagoras theorem

And the angle will be defined by:
tan(beta)=3/6
beta = tan^-1(6/3)
beta = 63.43°
Answer:
I like to memorize excerpt from articles to solve and answer questions like these. I hope this can help, it's from study.com: "The relationship between voltage, current, and resistance is described by Ohm's law. This equation, i = v/r, tells us that the current, i, flowing through a circuit is directly proportional to the voltage, v, and inversely proportional to the resistance, r."