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
A. because as the merry-go-round spins the child accelerates towards the center of the merry-go-round at a uniform rate.
Answer:
0.203 micro meter
Explanation:
for destructive interference that appearsblack, use the formula
2 t = m λ / u (where m = 0 1 2 3 ... is order of minima)
where t = tickness,
u is the ref index = 1.32
Wavelenth λ = 535×10^-9 meter
for t (minimum) m = 1 (as m=0 is ruled out as t>0)
t = 1× 535×10^-9/2×1.32
t (min) = 202.65×10^-9 meter
OR
t (min) = 0.203×10^-6 meter = 0.203 micro meter
Answer:
B
Explanation:
A body has kinetic energy that is moving
Answer:
Explanation:
m₂ is hanging vertically and m₁ is placed on inclined plane . Both are in limiting equilibrium so on m₁ , limiting friction will act in upward direction as it will tend to slip in downward direct . Tension in cord connecting the masses be T .
For equilibrium of m₁
m₁ g sinα= T + f where f is force of friction
m₁ g sinα= T + μsx m₁ g cosα
m₁ g sinα - μs x m₁ g cosα = T
For equilibrium of m₂
T = m₂g
Putting this value in equation above
m₁ g sinα - μs x m₁ g cosα = m₂g
m₂ = m₁ sinα - μs x m₁ cosα
