Answer:
a)
b)Does not affect the long term.
Explanation:
Given that

A = A0 cos(ωt)


This is linear equation so integration factor ,I


Now by using linear equation property



b)
at t= 0


So the initial condition does not affect the long term.
A Newton is a derived unit.
Hope that helped:)
Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
Answer:
Given
mass (m) =2kg
velocity (v) =3m/s
momentum (p) =?
Form
p=mv
2kgx3m/s
p=6kg.m/s
the momentum of ball's =6kg.m/s
Answer:
v = 3.04 m/s
Explanation:
given,
mass of the block, M = 6.6 Kg
horizontal force, F = 12.2 N
distance, L = 2.5 m
initial speed = 0 m/s
speed of the block,v = ?
we now
Work done is equal to change in Kinetic energy.
Work done = Force x displacement
W = Δ K E
Δ K E = Force x displacement


3.3 v² = 30.5
v² = 9.242
v = 3.04 m/s
speed of the block is equal to 3.04 m/s