Really? wow that's pretty cool
In this problem we have the electric field intensity E:
E = 6.5 × 
newtons/coulomb
We have the magnitude of the load:
q = 6.4 × 
coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × 
meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × 
)(6.5 × )(1.2 × )
PE = 5.0 x 
joules
None of the options shown is correct.
Answer:
Suppose that you are looking at the reflex candle in one mirror.
The image in the mirror is called the "virtual image"
If the real candle is at a distance of, let's say, 5 meters from the mirror, then the virtual image in the mirror will be also 5 meters "inside" the mirror.
Now, the light that emanates from the candle in the reflex seems to come from the virtual image, 5 meters inside of the mirror, but this is obviously not the real case.
The rays come, originally, from the real candle, and then those rays are reflected in the surface of the mirror.
Then the rays of the virtual image actually come from the surface of the mirror and not from the virtual image itself
Answer:
The acceleration of Mildred is, 1.33 m/s²
Explanation:
Given data,
The initial velocity of Mildred, u = 12 m/s
The final velocity of Mildred, v = 8 m/s
The time period of travel, t = 3 s
The acceleration of Mildred,
a = (v-u)/t
a = (8 - 12) /3
= -4/3
= -1.33 m/s²
The negative sign indicates that the Mildred is decelerating.
Hence, the acceleration of Mildred is, 1.33 m/s²
