I believe the correct answer from the choices listed above is the first option. Decreasing a telescope's eyepiece focal length will increase magnification. <span>The magnification of the </span>telescope<span> image is (</span>focal length<span> of the objective) divided by (</span>focal length <span>of the </span>eyepiece<span>). Hope this answers the question.</span>
Answer:
v = 16.11 m / s
Explanation:
For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation
starting point. When the spring is compressed
Em₀ = K_e + U = ½ k x² + m g x ’
final point. The point where it leaves the platform
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
½ k x² + m g x ’= ½ m v²
v² =
x² + g x
let's calculate
v² =
1.25² + 9.8 1.25
v² = 247.159 + 12.25 = 259.409
v = 16.11 m / s