Answer with Explanation:
We are given that


Charge on proton,q=
a.We have to find the electric potential of the proton at the position of the electron.
We know that the electric potential

Where 


B.Potential energy of electron,U=
Where
Charge on electron
=Charge on proton
Using the formula


Answer:
b. passes through the principal focal point.
Explanation:
Light wave can be defined as an electromagnetic wave that do not require a medium of propagation for it to travel through a vacuum of space where no particles exist.
A lens can be defined as a transparent optical instrument that refracts rays of light to produce a real image.
Basically, there are two (2) main types of lens and these includes;
I. Diverging (concave) lens.
II. Converging (convex) lens.
A converging lens refers to a type of lens that typically causes parallel rays of light with respect to its principal axis to come to a focus (converge) and form a real image. This type of lens is usually thin at the lower and upper edges and thick across the middle.
For a converging lens, a ray arriving parallel to the optic axis passes through the principal focal point.
Answer:
450X
Explanation:
When a specimen is been viewed, both
objective and ocular lenses works together so that the object is magnified.
From the question,objective lenses are;
1)10x
2)45x
ocular lens= 10x
Highest magnification
= 10X ocular × 45X objective
=450X
This implies that the image that was viewed will appear 450 times the actual size.
Answer:
Distance of 400m.
Explanation:
Use your kinematics equation to solve for distance (we can use kinematics b/c acceleration is constant).
d = (initial velocity x time) + 1/2 at^2
d = (20 x 10) + 1/2 (4) (10)^2
d = 200 + 200
d = 400 m
So power is equal to work over time and work is force times distance, you do 5 times 3 and get 15 dividing by 2 gives us 7.5 W answer c