According to the question, the object is placed at 2F
The ray diagram is shown in the figure attached.
According to the figure:
Object AB is at 2F₁
First, we draw a ray parallel to principal axis.
So, it passes through focus after refraction.
We draw another ray which passes through optical center.
So, the ray will go through without any deviation.
Where both refracted rays meet is point A' and the image formed is A'B'
This image is formed at 2F₂
We can say that:
- Image is real.
- Image is inverted.
- Image is exactly the same size as the object.
Answer:
<em>The total potential (magnitude only) is 11045.45 V</em>
Explanation:
<u>Electric Potential
</u>
The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by
Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:
The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical
The potential of the topmost charge is almost equal to the above computed, is only different in the sign:
The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':
The total electric potential in A is
The total potential (magnitude only) is 11045.45 V
Answer:
27.82998 km/min
Explanation:
To convert m/sec into km/hr, multiply the number by 18 and then divide it by 5.
