A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is connected to a movable object. The spring and object are compressed by 0.065 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 14.7 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.032 m relative to its unstrained length

0.832 m/s

Explanation:

The work done by the spring W equals the kinetic energy of the object K

The work done by the spring W = 1/2k(x₀² - x₁²) where k = spring constant, x₀   = initial compression = 0.065 m and x₁ = final compression = 0.032 m

The kinetic energy of the object, K = 1/2mv² where m = mass of object and v = speed of object

Since W = K,

1/2k(x₀² - x₁²) = 1/2mv²

k(x₀² - x₁²) = mv²

mv² = k(x₀² - x₁²)

v² = [(k/m)(x₀² - x₁²)]

taking square root of both sides, we have

v = √[(k/m)(x₀² - x₁²)] since ω = angular frequency = √(k/m),

v = √[(k/m)√(x₀² - x₁²)]

v = ω√(x₀² - x₁²)]

Since ω = 14.7 rad/s, we substitute the other variables into the equation, so we have

v = 14.7 rad/s × √((0.065 m)² - (0.032 m)²)]

v = 14.7 rad/s × √(0.004225 m² - 0.001024 m²)]

v = 14.7 rad/s × √(0.003201 m²)

v = 14.7 rad/s × 0.056577

v = 0.832 m/s

answer: d. a cme increases the amount of energy in the solar wind

explanation:

coronal mass ejections, also called cmes, are large clouds of plasma and magnetic field that erupt from the sun. only when the cloud is aimed at earth will the cme hit earth and therefore cause impacts. high-speed solar wind streams come from areas on the sun known as coronal holes.

power out to power in less than 100%