A mass is attached to a spring of spring constant 60 N/m along a horizontal, frictionless surface. The spring is initially stretched by a force of 5.0 N on the mass and let go. It takes the mass 0.50 s to go back to its equilibrium position when it is oscillating. What is the frequency of oscillation?

0.5 Hz

Explanation:

If it takes 0.5s to go back to the equilibrium position, that means it would take 1 s to reach maximum compression, and 2s to go back to the initial stretching position, aka complete 1 cycle of oscillation.

So the period is T = 2s. Then the frequency would be

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2