Answer:
The displacement in t = 0,
y (0) = - 0.18 m
Explanation:
Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N
v = √ T / μ
v = √20.48 N / 0.02 kg /m = 32 m/s
λ = v / f
λ = 32 m/s / 40 Hz = 0.8
K = 2 π / λ
K = 2π / 0.8 = 7.854
φ = X * 360 / λ
φ = 0.5 * 360 / 0.8 = 225 °
Using the model of y' displacement
y (t) = A* sin ( w * t - φ )
When t = 0
y (0) = 0.25 m *sin ( w*(0) - 225 )
y (0) = 0.25 * -0.707
y (0) = - 0.18 m
Answer:
(a) 1 : 2
(b) same
Explanation:
Let the mass of puck A is m and the mass of puck B is 2 m.
initial speed for both the pucks is same as u and the distance is same for both is s.
let the tension is T for same.
The kinetic energy is given by

(a) As the speed is same, so the kinetic energy depends on the mass.
So, kinetic energy of A : Kinetic energy of B = m : 2m = 1 : 2
(b) A the distance s same so the final velocities are also same.
The variation of water depth at spreading centers (ridges) controlled by isostasy as convective cooling cools the rocks much more effectively the than heat conduction.
<h3>What is convective heat transfer?</h3>
When heat transfer takes place between the two fluids in direct or indirect contact.
The lithosphere cools when it moves away from the ridge axis by sea floor spreading. The cooler rocks have low density, so the sea floor gets deeper as the lithosphere gets more dense.
Thus, the convective cooling cools the rocks much more effectively the than heat conduction.
Learn more about convective heat transfer
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Specific heat capacity= heat energy/mass×temperature rise
962°C - 20°C = 942K
Heat energy (Eh) = 239 × 1.55 × 942
Eh= 348963.9J
shc of Ag: 238.6 J/kg-K
m of Ag: 1.55kg
Explanation:
The particle will be at rest when its velocity
is equal to zero. Recall that the velocity is simply the derivative of the position
with respect to time:

Since 
then

Solving for t, we find that the particle will be at rest at
