When you heat something of cool it down you don't change the substance you might change the why is looks, but it is still the same substance. For example you cool water to 0 degrees Celsius it turns into ice but it still is two parts hydrogen and one part oxygen H2O. Physical changes will change state and/or form but it will still be what it originally was on the molecular level. Hope that helped.
Answer:
b) a = -k / m x
, c) d²x / dt² = - A w² cos (wt+Ф)
, d) and e) T = 2π √m / k
h) a = - A w² cos (wt+Ф)
Explanation:
a) see free body diagram in the attachment
b) We write Newton's second law
Fe = m a
-k x = ma
a = -k / m x
c) the acceleration is
a = d²x / dt²
If x = A cos wt
v = dx / dt = -A w sin (wt
+Ф)
a = d²x / dt² = - A w² cos (wt+Ф)
d) we substitute in Newton's second law
d²x / dt² = -k / m x
We call
w² = k / m
e) substitute to find w
-A w² cos (wt+Ф) = -k / m A cos (wt+Ф)
w² = k / m
Angular velocity and frequency are related
w = 2π f
f = 1 / T
We substitute
T = 2π / w
T = 2π √m / k
g) v= - A w sin (wt+Ф)
h) acceleration is
a = - A w² cos (wt+Ф)
To solve this problem we will apply the concepts related to energy conservation, so the potential energy in the package must be equivalent to its kinetic energy. From there we will find the speed of the package in the vertical component. The horizontal component is given, as it is the same as the one the plane is traveling to. Vectorially we will end up finding its magnitude. So,


Here,
m = Mass
g = Gravity
h = Height
v = Velocity
Rearranging to find the velocity

Replacing,


Using the vector properties the magnitude of the velocity vector would be given by,



Therefore the package is moving to 66.2m/s
Most of the excess energy is released as waste heat into the air surrounding the engine. Small amounts of excess energy are also released as sound energy, and as electrical energy generated by the alternator in a car's engine.