(h + .16) m g = 1/2 k x^2 total PE of block relative to where it stops
(h + .16) .82 * 9.8 = .5 * 120 * .16^2 PE released = PE of spring
8.04 h + 1.29 = 1.536
h = (1.536 - 1.29) / 8.04 = .031 m = 3.1 cm
Scientific form = 6.5 x 109.
Answer:
C. Quadruple the intensity
Explanation:
The intensity of the sound is proportional to square of amplitude of the sound.
I ∝ A²

When the given sound is twice loud as the initial value, then the new amplitude is twice the former.
A₂ = 2A₁

Thus, to make a given sound seem twice as loud, the musician should Quadruple the intensity
When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is
∑ F = R - mg = 0
where mg = weight of the mass = (7.00 kg) g = 68.6 N.
It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that
k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m