Answer:
The answer is "512 J".
Explanation:
bullet mass 
initial speed 
block mass
initial speed
final speed 
Let
will be the bullet speed after collision:
throughout the consevation the linear moemuntum
The kinetic energy of the bullet in its emerges from the block


By Newton's second law, the net vertical force acting on the object is 0, so that
<em>n</em> - <em>w</em> = 0
where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that
80 N = <em>µ</em> (196 N) → <em>µ</em> = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is
40 N = <em>ν</em> (196 N) → <em>ν</em> = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)
In spring mass system we know that angular frequency is given as

f = 8.38 Hz


now we know that speed of SHM at its extreme position is given by

here we know that
A = 17.5 cm


so maximum speed is 9.21 m/s
Answer:
The helicopter was 1103.63 meters high when the package was dropped.
Explanation:
We consider positive speed as a downward movement
y: height (m)
t: time (s)
v₀: initial speed (m/s)
Δy = v₀t +
gt²
Δy= 15
×15 s +
×9.81
×(15 s)²
Δy= 1103.63 m