Answer:
The back end of the vessel will pass the pier at 4.83 m/s
Explanation:
This is purely a kinetics question (assuming we're ignoring drag and other forces) so the weight of the boat doesn't matter. We're given:
Δd = 315.5 m
vi = 2.10 m/s
a = 0.03 m/s^2
vf = ?
The kinetics equation that incorporates all these variables is:
vf^2 = vi^2 + 2aΔd
vf = √(2.1^2 + 2(0.03)(315.5))
vf = 4.83 m/s
Transverse Waves will be your answer I just saw the answer
Answer:
0.069 J
Explanation:
mass of first ball (M1) = 0.839 kg
mass of second ball (M2) = 0.839 kg
initial velocity of first ball (U1) = 0.406 m/s
initial velocity of second ball (U2) = 0 m/s
For head on elastic collisions of equal masses, the velocities of the masses always changes.and this has also been clearly stated in the question. Hence the velocity if mass A will interchange with that of mass B after collision.
therefore
final velocity of first ball (V1) = 0 m/s
final velocity of second ball (V2) = 0.406 m/s
kinetic energy of second ball after collision = 0.5m
= 0.5 x 0.839 x 0.406 x 0.406 = 0.069 J
Answer:
Part 1: 0.3789
Part 2: 746 J
Part 3: 2.162 kW
Explanation:
Part 1:
Eff= 
Eff= 0.378873 ≈ 0/3789
Part 2:
W= 0.3789(1969)
W= 746 J
Part 3:

Power= 2162.3188 Watts
2162.3188 W-----> 2.162 kW