Answer:
final velocity will be44.72m/s
Explanation:
HEIGHT=h=100m
vi=0m/s
vf=?
g=10m/s²
by using third equation of motion for bodies under gravity
2gh=(vf)²-(vi)²
evaluating the formula
2(10m/s²)(100m)=vf²-(0m/s)²
2000m²/s²=vf²
√2000m²/s²=√vf²
44.72m/s=vf
Answer:The frequency of the echo is slightly decreased
Explanation:
Given
speed of boat 
cliff is 
away
when boat is still , suppose t is the time taken by the echo to reach observer on the boat
But as soon as boat starts moving the distance between cliff and boat decreasing and time for echo to reach observer also decreases
and we know 
therefore frequency of the echo slightly decreased.
Answer:
v = 2.94 m/s
Explanation:
When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.
Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.
Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means
(1/2)kx^2 = (1/2)mv^2
kx^2 = mv^2
v^2 = (kx^2)/m
v = sqrt((kx^2)/m)
v = x * sqrt(k/m)
v = 0.122 * sqrt(125/0.215) <--- units converted to m and kg
v = 2.94 m/s
