Cocking your head would be most useful for detecting the LOCATION of a sound.
Answer:

ΔK = 2.45 J
Explanation:
a) Using the law of the conservation of the linear momentum:

Where:


Now:

Where
is the mass of the car,
is the initial velocity of the car,
is the mass of train,
is the final velocity of the car and
is the final velocity of the train.
Replacing data:

Solving for
:

Changed to cm/s, we get:

b) The kinetic energy K is calculated as:
K = 
where M is the mass and V is the velocity.
So, the initial K is:



And the final K is:




Finally, the change in the total kinetic energy is:
ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J
This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.
Answer:
1.02 m/s²
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 0 m/s
Final velocity (v) = 6.6 m/s
Time (t) = 6.5 s
Acceleration (a) =.?
Acceleration can simply be defined as the change of velocity with time. Mathematically, it can be expressed as:
a = (v – u) / t
Where:
a is the acceleration.
v is the final velocity.
u is the initial velocity.
t is the time.
With the above formula, we can obtain the acceleration of the car as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 6.6 m/s
Time (t) = 6.5 s
Acceleration (a) =.?
a = (v – u) / t
a = (6.6 – 0) / 6.5
a = 6.6 / 6.5
a = 1.02 m/s²
Therefore, the acceleration of the car is 1.02 m/s²