Answer:
The Magnifying power of a telescope is
Explanation:
Radius of curvature R = 5.9 m = 590 cm
focal length of objective =
⇒ =
⇒ = 295 cm
Focal length of eyepiece = 2.7 cm
Magnifying power of a telescope is given by,
therefore the Magnifying power of a telescope is
Answer:
Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e ()
with this expression we can find the closest approach distance (r)