When an open cylindrical tank, with a diameter of 2 meters and a height of 4 meters, is rotated about its vertical axis at a constant angular speed of 90 rpm, the amount of water spilled can be determined by calculating the volume of the spilled water.
By considering the geometry of the tank and the rotation speed, the spilled water volume can be calculated. The calculation involves finding the height of the water level when rotating at the given angular speed and then calculating the corresponding volume. The answer to the question is the option that represents the calculated volume in liters.
To determine the amount of water spilled, we need to calculate the volume of the water that extends above the half-full level of the cylindrical tank when it is rotated at 90 rpm.First, we find the height of the water level at the given angular speed. Since the tank is half-full, the water level will form a parabolic shape due to the centrifugal force. The height of the water level can be calculated using the equation h = (1/2) * R * ω^2, where R is the radius of the tank (1 meter) and ω is the angular speed in radians per second.
Converting the angular speed from rpm to radians per second, we have ω = (90 rpm) * (2π rad/1 min) * (1 min/60 sec) = 3π rad/sec. Substituting the values into the equation, we find h = (1/2) * (1 meter) * (3π rad/sec)^2 = (9/2)π meters. The height of the spilled water is the difference between the actual water level (4 meters) and the calculated height (9/2)π meters. Therefore, the height of the spilled water is (4 - (9/2)π) meters.
To find the volume of the spilled water, we calculate the volume of the frustum of a cone, which is given by V = (1/3) * π * (R1^2 + R1 * R2 + R2^2) * h, where R1 and R2 are the radii of the top and bottom bases of the frustum, respectively, and h is the height. Substituting the values, we have V = (1/3) * π * (1 meter)^2 * [(1 meter)^2 + (1 meter) * (1/2)π + (1/2)π^2] * [(4 - (9/2)π) meters].
By evaluating the expression, we find the volume of the spilled water. To convert it to liters, we multiply by 1000. The option that represents the calculated volume in liters is the correct answer. Answer is d. 768
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Show that the free-particle one-dimensional Schro¨dinger
equation for the wavefunc-
tion Ψ(x, t):
∂Ψ
i~
∂t = −
~
2
2m
∂
2Ψ
,
∂x2
is invariant under Galilean transformations
x
′ = x −
3. Galilean invariance of the free Schrodinger equation. (15 points) Show that the free-particle one-dimensional Schrödinger equation for the wavefunc- tion V (x, t): at h2 32 V ih- at is invariant u
The Galilean transformations are a set of equations that describe the relationship between the space-time coordinates of two reference systems that move uniformly relative to one another with a constant velocity. The aim of this question is to demonstrate that the free-particle one-dimensional Schrodinger equation for the wave function ψ(x, t) is invariant under Galilean transformations.
The free-particle one-dimensional Schrodinger equation for the wave function ψ(x, t) is represented as:$$\frac{\partial \psi}{\partial t} = \frac{-\hbar}{2m} \frac{\partial^2 \psi}{\partial x^2}$$Galilean transformation can be represented as:$$x' = x-vt$$where x is the position, t is the time, x' is the new position after the transformation, and v is the velocity of the reference system.
Applying the Galilean transformation in the Schrodinger equation we have:
[tex]$$\frac{\partial \psi}{\partial t}[/tex]
=[tex]\frac{\partial x}{\partial t} \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial t}$$$$[/tex]
=[tex]\frac{-\hbar}{2m} \frac{\partial^2 \psi}{\partial x^2}$$[/tex]
Substituting $x'
= [tex]x-vt$ in the equation we get:$$\frac{\partial \psi}{\partial t}[/tex]
= [tex]\frac{\partial}{\partial t} \psi(x-vt, t)$$$$\frac{\partial \psi}{\partial x} = \frac{\partial}{\partial x} \psi(x-vt, t)$$$$\frac{\partial^2 \psi}{\partial x^2} = \frac{\partial^2}{\partial x^2} \psi(x-vt, t)$$[/tex]
Substituting the above equations in the Schrodinger equation, we have:
[tex]$$\frac{\partial}{\partial t} \psi(x-vt, t) = \frac{-\hbar}{2m} \frac{\partial^2}{\partial x^2} \psi(x-vt, t)$$[/tex]
This shows that the free-particle one-dimensional Schrodinger equation is invariant under Galilean transformations. Therefore, we can conclude that the Schrodinger equation obeys the laws of Galilean invariance.
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