Asolid copper cube is attached to spring-like device and is able to oscillate horizontally with negligible friction. each side of the cube is 1.50 cm long. the cube is initially pulled to a point where it stretches the spring-like device by 2.65 cm where it is held at rest with a horizontal force of 1.50 n. what is the frequency at which it oscillates when it is released? (assume the density of copper is 8.92 g/cm3.)

6.91 Hz

Explanation:

Volume of the cube

= (1.5 x 10⁻² )³m³

= 3.375 x 10⁻⁶ m³

mass of the cube

= 3.375 x 10⁻⁶  x 8920 ( 8.92 g / cm³ = 8920 kg/m³ )

m = 30.105 x 10⁻³ kg

Spring stretches by 2.65 x 10⁻² m due to a force of  1.5 N.

If k be the force constant

k x = F

K x2.65 x 10⁻²  = 1.5

k = .566 x 10² N / m

Now frequency of oscillation for spring - mass system is given by

n =

=

n = 6.91 Hz

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

higher potential

explanation:

because it is in a high speed so it must move into a of high potential