I'm not that smart but I think it is c I really hope It helps
Answer:
The equation for the object's displacement is 
Explanation:
Given:
m = 16 lb
δ = 3 in
The stiffness is:

The angular speed is:

The damping force is:

Where
FD = 20 lb
u = 4 ft/s = 48 in/s
Replacing:

The critical damping is equal:

Like cc>c the system is undamped
The equilibrium expression is:

It makes no sense how you typed this problem out.
Answer:
The (s) indicates that the state of matter for NaHCO3 is solid.
Explanation:
When a chemical reaction is written, the state of matter for each components of the reactants and products are mentioned in brackets along with their names or formulas.
For example, NaHCO3 has (s) mentioned in the brackets. The s shows that the state of matter for NaHCO3. (l) represents liquid format. (g) represents that the state of matter is gas.
Answer:
D. the proper replacement unit for one joule per second
Explanation:
When energy is divided by the time the energy was used we get power



So, the answer is D. the proper replacement unit for one joule per second