D (Glucose +Oxygen --> Carbon Dioxide + Water + Energy)
Answer:
the inner core if I believe. I apologize if I am incorrect
Answer:
3.9 m/s
Explanation:
We are given that
Mass of car,m=
Initial velocity,u=0
Distance,s=5.9 m

Average friction force,f=
We have to find the speed of the car at the bottom of the driveway.
Net force,
Where 
Acceleration,


v=3.9 m/s
Answer:
Part a)
A = 0.066 m
Part b)
maximum speed = 0.58 m/s
Explanation:
As we know that angular frequency of spring block system is given as

here we know
m = 3.5 kg
k = 270 N/m
now we have


Part a)
Speed of SHM at distance x = 0.020 m from its equilibrium position is given as



Part b)
Maximum speed of SHM at its mean position is given as

