Two friends are playing a version of proton golf where the hole is marked by a single

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

     = (9×10^9)×(q)/(24.0)

     = 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

                       R^2 = 9×r^2

E2 = k×q/(9×r^2)

     = k×q/(9×375000000q)

     = k/(9×375000000)

     = (9×10^9)/(9×375000000)

     = 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

 f = - 20 cm

Explanation:

This exercise asks us for the focal length, which for a lens in air is

                  1 / f = (n₂-n₁) (1 / R₁ - 1 / R₂)

where n₂ is the refractive index of the material, n₁ is the refractive index of the medium surrounding the lens, R₁ and R₂ are the radii of the two surfaces.

In this exercise the medium that surrounds the lens is air n₁ = 1 and the lens material has an index of refraction n₂ = n = 1.50, let's substitute in the expression

                 - 1/40 = (n-1) (1 / R₁ -1 / R₂)

                (1 / R₁ - 1 / R₂) = - 1/40 (n-1)

let's calculate

               (1 / R₁ -1 / R₂) = - 1/40 (1.50 -1)

               (1 / R₁ -1 / R₂) = -1/20

 Now we change the construction material for one with refractive index

n = 2, keeping the radii,

              1 / f = (n-1) (1 / R₁-1 / R₂)

              1 / f = (n-1) (-1/20)

let's calculate

             1 / f = (2.00-1) (-1/20)

              1 / f = -1/20

              f = - 20 cm