I'm not 100% sure, but I believe what you mean is when they eject the old propulsion motors. Yes, they land in the ocean and the US Navy retrieves them for later use.
Answer:
Explanation:
Total momentum of the system before the collision
.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left
If v be the velocity of the stuck pucks
momentum after the collision = 2 v
Applying conservation of momentum
2 v = - .75
v = - .375 m /s
Let after the collision v be the velocity of .5 kg puck
total momentum after the collision
.5 v + 1.5 x .231 = .5v +.3465
Applying conservation of momentum law
.5 v +.3465 = - .75
v = - 2.193 m/s
2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.
Kinetic energy before the collision
= 2.25 + 1.6875
=3.9375 J
kinetic energy after the collision
= .04 + 1.2 =1.24 J
So kinetic energy is not conserved . Hence collision is not elastic.
3 ) Change in the momentum of .5 kg
1.5 - (-1.0965 )
= 2.5965
Average force applied = change in momentum / time
= 2.5965 / 25 x 10⁻³
= 103.86 N
is the horse running/moving? if so then kinetic i believe
Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
U=0
<span>t=10 </span>
<span>a=9.8m/s/s </span>
<span>v is velocity (the tower must be very high to be able to fall for 10 seconds!!!) </span>
<span>you work out the result now</span>
