A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled has an acceleration of 2.10 m/s2. Part A By how much does the spring stretch if it pulls on the sled horizontally

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring  (T) = ma

Tension force in the spring  (T) = 9 × 2.10

Tension force in the spring  (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N  × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

divide the right side by the left side

0.024÷1.2=0.02 for example

layers of the earth affect the waves and these effects are seen on instruiments such as seimometers