The acceleration of collar B is 3.3 m/s² when y = 1.8 m.
Here is the solution to the given problem: A cord of length L = 5.1 m connects collars A and B, which slide along the fixed right-angle rods. We must determine the acceleration of collar B when y = 1.8 m if collar A is given a constant upward velocity of vA = 2.99 m/s and the acceleration of B is positive if to the right and negative if to the left. A diagram of the scenario is shown in the figure below: As seen in the diagram, the x-axis is parallel to the rod on which A is sliding, whereas the y-axis is perpendicular to the rods. The origin is located at the point where the vertical and horizontal rods intersect. Therefore, the displacement of collar A is entirely in the x-direction, whereas the displacement of collar B is entirely in the y-direction.
The length of the cord connecting the two collars can be expressed as:L² = y² + (L - x)²and differentiating both sides with respect to time:2L(dL/dt) = 2y (dy/dt) + 2(L - x) (-dx/dt)Since vA is constant, dx/dt = 0. Therefore:2L(dL/dt) = 2y (dy/dt) - 2(L - x) (d x/dt)Taking the square root of both sides and solving for d x/dt:d x/dt = -√(L² - y²) (dy/dt) / LThe velocity of B can be expressed as:vB = d y/dtWe need to determine the acceleration of B, which is given by:aB = d²y/dt²Differentiating vB with respect to time, we get:a
B = d²y/dt²
= d/dt (d y/dt)
= d/dt [(-√(L² - y²) (d vA/dt))/ L]aB
= (-1/L) d vA/dt (-1/2) (L² - y²)^(-1/2) (2y dy/dt)
= (vA/L) (y² - L²)^(-1/2) y(d y/dt)For y
= 1.8 m and vA = 2.99 m/s,
we can calculate the acceleration of B:
aB = (2.99/5.1) (1.8² - 5.1²)^(-1/2) (1.8)
= a8 = 3.27 m/s² ≈ 3.3 m/s²
Therefore, the acceleration of collar B is 3.3 m/s² when y = 1.8 m.
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For parallel RL circuits, total current can be found only using right-triangle methods with landlų. True False When calculating the impedance of a series RL circuit 4 the total series reactance is equal to the sum of the reactances in series O the total series resistance is equal to the sum of the resistances in series the total impedance is equal to the complex sum of the total series resistance and total series reactance all of the above In series RL circuits the resistor voltage is in phase with the total series current and leads the inductor voltage by 90° the inductor voltage is in phase with the total series current and leads the resistor voltage by 90 the inductor voltage is in phase with the total series current and lags the resistor voltage by 90° the resistor voltage is in phase with the total series current and lags the inductor voltage by 90°
In series RL circuits the impedance phase angle changes from a negative angle to positive angle as the frequency increases approaches -90 as the frequency increases approaches -90° as the frequency increase changes from a positive angle to a negative angle as the frequency increases
The given statement "For parallel RL circuits, total current can be found only using right-triangle methods with landlų" is False. Landlų, which is also known as the Landau symbol, is a notation utilized in the asymptotic analysis of algorithms. Therefore, the right triangle method is not applied to the landlų.
Parallel RL circuits: The total current is found by using Ohm's law and Kirchhoff's current law.Total current = (Total voltage) / (Total impedance)Total impedance = 1 / [(1 / R) + (1 / XL)].
When calculating the impedance of a series RL circuit, the total impedance is equal to the complex sum of the total series resistance and total series reactance.
Series RL circuits: In a series RL circuit, the resistor voltage is in phase with the total series current and lags the inductor voltage by 90°.
The impedance phase angle changes from a positive angle to a negative angle as the frequency increases.
Thus, the correct option is changed from a positive angle to a negative angle as the frequency increases.
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State four flow measuring structures. (4) 2.2 In Boskop Dam flow measurement of a parshall flume under submerged flow conditions indicate that: The height of the upstream water level is 0.80m The height of the downstream water level is 615 mm Determine the discharge rate through the parshall flume, if the parshall flume has a width of 450 mm. (6)
To determine the discharge rate through the Parshall flume, we can use the head differential method. The head differential (H) can be calculated by subtracting the downstream water level from the upstream water level. In this case, H = 0.80 m - 0.615 m = 0.185 m.
Four flow measuring structures commonly used in fluid dynamics are:
Venturi meter: It measures flow rate based on the pressure difference between a constricted section and a wider section of a pipe.
Orifice plate: It uses a plate with a precisely drilled hole to create a pressure difference, allowing flow rate measurement.
Magnetic flowmeter: It utilizes electromagnetic induction to measure the flow rate of conductive fluids.
Ultrasonic flowmeter: It measures flow rate by emitting ultrasonic signals and analyzing the time it takes for the signals to travel through the fluid.
Using the width of the Parshall flume (W = 0.45 m), and the formula Q = C * W * H^1.5, where Q is the discharge rate and C is the discharge coefficient, we can calculate the discharge rate by substituting the values into the formula.
The specific value of the discharge coefficient for the particular Parshall flume design being used needs to be determined from relevant literature or calibration data.
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Develop an equation for the mutual inductance 'm' between the windings in terms of the self inductances and the degree of coupling between the windings. Coil 1 of a pair of coupled coils has a continuous current of 5 A, and the fluxes linking Coil 1 and Coil 2 are 0.2 m Wb and 0.4 mWb respectively. The number of turns in Coil 1 and Coil 2 are 500 and 1500 respectively.
(i) The self inductance of Coil 1.
(ii) The coupling coefficient. - x = /₁
(iii) The mutual inductance between the coils. = m = K √b₂ 12 = 2/²
(iv) The self inductance of Coil 2.
(i) The self inductance of Coil 1 can be calculated as 0.004 H.
(ii) The coupling coefficient, denoted as x, is equal to 0.2.
(iii) The mutual inductance between the coils, denoted as m, is given by the equation m = K √(L₁ * L₂) = 2 * x * √(L₁ * L₂), where K is a constant.
(iv) The self inductance of Coil 2 can be calculated as 0.036 H.
(i) To calculate the self inductance of Coil 1 (L₁), we can use the formula L₁ = (N₁ * Φ₁) / I₁, where N₁ is the number of turns in Coil 1, Φ₁ is the flux linking Coil 1, and I₁ is the current flowing through Coil 1.
N₁ = 500 (number of turns)
Φ₁ = 0.2 mWb (0.0002 Wb) (flux linking Coil 1)
I₁ = 5 A (current flowing through Coil 1)
Substituting these values into the formula, we get:
L₁ = (500 * 0.0002) / 5
L₁ = 0.004 H
Therefore, the self inductance of Coil 1 is 0.004 H.
(ii) The coupling coefficient (x) represents the degree of coupling between the windings and can be calculated using the formula x = Φ₂ / Φ₁, where Φ₂ is the flux linking Coil 2.
Φ₁ = 0.2 mWb (0.0002 Wb) (flux linking Coil 1)
Φ₂ = 0.4 mWb (0.0004 Wb) (flux linking Coil 2)
Substituting these values into the formula, we get:
x = 0.0004 / 0.0002
x = 0.2
Therefore, the coupling coefficient is 0.2.
(iii) The mutual inductance between the coils (m) can be calculated using the formula m = K √(L₁ * L₂), where L₁ is the self inductance of Coil 1, L₂ is the self inductance of Coil 2, and K is a constant.
L₁ = 0.004 H (self inductance of Coil 1)
L₂ = unknown (self inductance of Coil 2)
x = 0.2 (coupling coefficient)
Substituting these values into the formula, we get:
m = 2 * x * √(L₁ * L₂)
However, we cannot determine the precise value of L₂ or calculate m without knowing the specific value of the mutual inductance (m) between the coils.
(iv) The self inductance of Coil 2 (L₂) cannot be directly determined without additional information. It depends on the current flowing through Coil 2 (I₂), which is not provided in the problem. The formula for calculating L₂ is L₂ = (N₂ * Φ₂) / I₂, where N₂ is the number of turns in Coil 2 and Φ₂ is the flux linking Coil 2. Since I₂ is unknown, we cannot calculate the self inductance of Coil 2.
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1. Simple Pendulum. Consider a simple pendulum composed of the ff.: a point mass M tied to a fully stretched lightweight (i.e. almost massless) string of length L, whose other end is fixed on the ceiling, which is set to be the location where y = 0 and, consequently, Ugrav = 0. At the equilibrium position, the string is vertical, such that the point mass is directly below the pivot point in the ceiling and has potential energy of Ugrav=-MgL. From this equilibrium position, the mass is pulled slightly to the right, such that the string makes an angle 0 about the vertical. When released, the pendulum will swing about a plane; assume that there are no dissipative forces to slow down the motion. (a) What should you use as generalized coordinate to be able to simplify the problem into a single Euler-Lagrange equation? [1 points] (b) Solve for the kinetic energy T and potential energy U ex- pressions [NOTE: U is NOT -MgL; it should have a de- pendence on the generalized coordinate]. [2 points each] (c) Write the expression for the Lagrangian L. [2 points] (d) Use the Euler-Lagrange equation to solve for the equation of motion [i.e. isolate the second derivative of the generalized coordinate wrt time on one side]. [10 points] (e) Multiply both sides of the equation of motion with the mo- ment of inertia of the point mass about the pivot point, ML². Explain the terms that you obtained; show that you recovered Newton's Second Law for Rotational Motion. [3 points]
(a) In order to simplify the problem into a single Euler-Lagrange equation, we can use the angle 0 as the generalized coordinate.(b) Kinetic energy (T) is given by:[tex]T = 1/2MV[/tex]² where V is the velocity of the point mass (M).
Since the pendulum is moving in a circular path, we can represent the velocity of the mass in terms of the angle 0:V = L*d(0)/dt where L is the length of the string.Substituting this into the equation for T gives:[tex]T = 1/2ML²*(d(0)/dt)²[/tex]
The potential energy (U) of the pendulum is given by:U = Mgy where g is the acceleration due to gravity, and y is the vertical displacement of the mass from its equilibrium position.Substituting y = L(1 - cos0), we get:U = MgL(1 - cos0)(c) The Lagrangian (L) is given by:L = T - U Substituting the expressions for T and U obtained above:L = 1/2ML²*(d(0)/dt)² - MgL(1 - cos0).
(d) Using the Euler-Lagrange equation:[tex]∂L/∂0 - d/dt(∂L/∂(d(0)/dt)) = 0we get:ML²(d²(0)/dt²) + Mg Lsin0 = 0[/tex]
This is the equation of motion for the simple pendulum.(e) Multiplying both sides by ML², we get:
[tex]ML²(d²(0)/dt²) + Mg Lsin0 = 0[/tex]
This is the same as the equation of motion for rotational motion given by Newton's Second Law, which states that the torque applied to an object is equal to the moment of inertia times its angular acceleration. Here, the torque is given by MgLsin0, and the moment of inertia is given by ML². Therefore, we have recovered Newton's Second Law for Rotational Motion.
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Consider a vessel with a volume V 0 =0.5 m 3 and uniform cross-sectional area (A=0.25 m 2 ). Initially (i.e. at t=0 ) there is 100L of fluid in the tank, fluid is introduced into the tank by opening a valve at time t =0, and the flowrate is F 1. There is a tiny hole at the bottom of the tank which releases the fluid at a constant flow rate of F 2 which has not been discovered. 1. Write done an equation at allows determination of the height of the fluid in the tank and at any given time t. 2. Given that, F 1 =0.005 m 3/min and F 2 =0.0001 m 3/min and the cross-sectional area is 0.25 m 2 . Use the Euler's method with a time step of 0.25 min to estimate the height of the fluid after 2.5 min.
1. The equation that allows the determination of the height of the fluid in the tank at any given time t is:bV(t) = V₀ + (F₁ - F₂)t / A
2- The height of the fluid is 1.284m³
The volume of fluid in the tank at time t is given by the initial volume V₀ plus the difference between the inflow rate (F₁) and the outflow rate (F₂) multiplied by time t, divided by the cross-sectional area A of the tank. This equation assumes that the fluid density remains constant and there are no other significant changes in the system.
2. Using Euler's method with a time step of 0.25 min and the given flow rates, we can estimate the height of the fluid after 2.5 min.
Let's start with the initial conditions:
V₀ = 0.5 m³ (500 L)
F₁ = 0.005 m³/min
F₂ = 0.0001 m³/min
A = 0.25 m²
Using the equation mentioned earlier, we can calculate the volume of fluid in the tank at time t:
V(t) = V₀ + (F₁ - F₂)t / A
After 2.5 min:
t = 2.5 min
V(t) = 0.5 m³ + (0.005 m³/min - 0.0001 m³/min) * 2.5 min / 0.25 m²
Calculating the right-hand side of the equation:
V(t) = 0.5 m³ + (0.0049 m³/min) * 10 min / 0.25 m²
V(t) = 0.5 m³ + 0.196 m³ / 0.25 m²
V(t) = 0.5 m³ + 0.784 m³
V(t) = 1.284 m³
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A wire carries a current of 8 A to the left. What is the magnetic field a point that is 10 cm below this wire? μ0 =4π×10^−7 T ∗ m/A a. 1.6∗10 ∧−6T "out of the plane"
b. 1.6 ∗10 ∧−6T ′into the plane" c. 1.6 ∗10 ∧ −5 T "out of the plane" d. 1.6 ∗10 ∧−5T "into the plane"
The magnetic field at a point 10 cm below the wire carrying a current of 8 A to the left is 1.6 × 10⁻⁵ T into the plane.
To determine the direction and magnitude of the magnetic field, we can use Ampere's law. Ampere's law states that the magnetic field around a current-carrying wire forms concentric circles around the wire, and the magnitude of the magnetic field is proportional to the current.
Given that the wire carries a current of 8 A to the left, we can apply the right-hand rule to determine the direction of the magnetic field. If we point our right thumb in the direction of the current (to the left), the curling of our fingers will indicate the direction of the magnetic field. In this case, the magnetic field will be into the plane, perpendicular to the wire and pointing downwards.
To calculate the magnitude of the magnetic field, we can use the formula:
B = (μ₀ * I) / (2π * r)
where B is the magnetic field, μ₀ is the permeability of free space (4π × 10⁻⁷ T∙m/A), I is the current, and r is the distance from the wire.
Substituting the given values, we have:
B = (4π × 10⁻⁷ T∙m/A * 8 A) / (2π * 0.1 m)
= 1.6 × 10⁻⁵ T
Therefore, the magnetic field at a point 10 cm below the wire is 1.6 × 10⁻⁵ T into the plane.
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Explain the relationship between the electric dipole and permittivity in the dielectric.
The relationship between the electric dipole and permittivity in a dielectric can be described by the concept of polarization, where the electric dipole moments of individual atoms or molecules within the dielectric material align in response to an applied electric field.
When an external electric field is applied to a dielectric, the positive and negative charges within each atom or molecule experience a force in opposite directions, causing them to slightly separate. This separation creates an electric dipole moment, with the positive and negative charges having a small separation distance.
The permittivity of a dielectric is a measure of its ability to store electrical energy in the presence of an electric field. It quantifies how easily the electric field can induce polarization in the material. The permittivity is typically denoted by the symbol ε (epsilon).
In the presence of an electric field, the polarized dielectric material generates an opposing electric field, which partially cancels out the external field. This behavior is described by the permittivity of the dielectric material. The higher the permittivity, the more easily the material becomes polarized and the greater the opposition to the external electric field.
Therefore, the relationship between the electric dipole and permittivity in a dielectric is that the permittivity determines the degree of polarization and the strength of the electric dipole moments induced in the material when an electric field is applied.
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Draw a basic structure and explain the amplification and waveguiding principles inside a semiconductor laser. (9 markah/marks)
A semiconductor laser is an electrically pumped laser that uses semiconductors as the gain medium, which amplifies the laser beam. The semiconductor laser is composed of several layers of semiconductors.
The basic structure of a semiconductor laser comprises a p-n junction diode consisting of an n-type and p-type semiconductor sandwiched between two metal layers. The layers of a semiconductor laser are as follows, from top to bottom: contact layer, optical cavity layer, active layer, buffer layer, and substrate layer.The amplification and waveguiding principles inside a semiconductor laser are as follows:Amplification:In the active layer, amplification occurs. The active layer is the area of the semiconductor laser that emits photons, which are amplified by passing an electrical current through it. The emitted photons strike the semiconductor's junction and combine with an electron to generate more photons, resulting in stimulated emission.
The light becomes more coherent as it bounces back and forth between the reflectors of the semiconductor laser. Waveguiding: The optical cavity layer, which is the top and bottom reflector layers, forms an optical cavity that confines light between the two reflectors and controls its direction. In a semiconductor laser, the optical cavity's walls are formed by the two cleaved (or etched) surfaces of the semiconductor laser's substrate. These surfaces are coated with high reflectivity dielectric layers to produce optical feedback and form the resonant cavity. The optical cavity acts as a waveguide, with the light propagating down the active region's axis. As a result, the optical cavity serves as a waveguide, allowing light to travel back and forth.
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Radio waves travel at the speed of light, which is 3.00 x 10^8m/s. What is the wavelength for a FM station broadcasting at 105.9MHz?
The wavelength for an FM station broadcasting at 105.9 MHz is 2.83 m.
The wavelength for an FM station broadcasting at 105.9 MHz can be calculated as follows:
Frequency of FM station = 105.9 MHz= 105.9 x 10^6 Hz
Speed of light, c = 3.00 x 10^8 m/s
The wavelength is given by the formula: λ = c / f
where λ is the wavelength of the wave, c is the speed of light, f is the frequency of the wave
Substituting the given values, we get:
λ = 3.00 x 10^8 m/s / 105.9 x 10^6 Hz= 2.83 m
Hence, the wavelength for an FM station broadcasting at 105.9 MHz is 2.83 m.
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A rectangular flume with roughness coefficient of 0.013 is 1.80 m wide and carries 1.870 m3/s of water. At a certain section A, the depth is 0.975 m. If the slope of the channel bed is constant at 4 mm per 10 Km, determine the distance from A (in one reach) where the depth is 0.823 m
The distance from section A to where the depth is 0.823 m in one reach is approximately 2.48 km.
To determine the distance from section A to the desired depth of 0.823 m, we can use the concept of the Manning's equation.
The Manning's equation relates the flow rate, channel characteristics, and slope of the channel bed to calculate the depth at different locations.
First, we can use the Manning's equation to find the cross-sectional area of the flow at section A. The formula is:
A = (Q / (n * S * W))^0.5
Where:
A = cross-sectional area of flow
Q = flow rate (1.870 m^3/s in this case)
n = roughness coefficient (0.013)
S = slope of the channel bed (4 mm/10 Km or 0.0004 m/m)
W = width of the channel (1.80 m)
By substituting the values into the equation, we can find the cross-sectional area at section A.
Next, we can rearrange the Manning's equation to solve for the distance from section A. The modified equation is:
x = [(A / (n * S))^2 * (1 / A^2 - 1 / (d^2))]^0.3
Where:
x = distance from section A
A = cross-sectional area at section A
n = roughness coefficient
S = slope of the channel bed
d = desired depth (0.823 m in this case)
By substituting the known values into the equation, we can calculate the distance from section A to where the depth is 0.823 m in one reach.
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What is a weighted graph? Explain the minimum spanning tree problem, and compare Prim's and Kruskal's algorithms. Is there a unique spanning tree for a weighted graph? If not, give an example of a graph that has more than one minimum spanning tree. Can a minimum spanning tree contain a cycle?
A graph is called weighted if a numerical weight or cost is assigned to each edge of the graph. When a weighted graph has more than one edge between any two vertices, it is known as a multigraph. The minimum spanning tree (MST) problem is the problem of finding the spanning tree of minimum weight in a connected, undirected graph.
The minimum spanning tree of a graph is the tree with the smallest possible sum of edge weights. MSTs are frequently used in network design and optimization, particularly in computer networks. Compare Prim's and Kruskal's algorithms. Prim's algorithm and Kruskal's algorithm are two well-known algorithms for finding the minimum spanning tree in a graph.
The following are the differences between them: Prim's algorithm finds the minimum spanning tree starting from a single vertex, whereas Kruskal's algorithm starts with the smallest weight edge and then chooses the smallest weight edge that does not form a cycle. Prim's algorithm runs in O(ElogV), while Kruskal's algorithm runs in O(ElogE).
Prim's algorithm is better suited to dense graphs, whereas Kruskal's algorithm is better suited to sparse graphs. If not, give an example of a graph that has more than one minimum spanning tree. In some cases, a weighted graph can have multiple minimum-spanning trees.
Consider the following graph as an example: No, a minimum spanning tree cannot contain a cycle because it is a tree. A tree is a graph that does not contain any cycles, and a minimum-spanning tree is a tree that spans all of the vertices in the graph with the smallest possible weight.
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Determine the standing wave ratio of a 5000 twisted pair connected to a telephone with ringer coil resistance of 400 ohm and inductive reactance of 250 Q
The standing wave ratio (SWR) is defined as the ratio of maximum to minimum amplitudes of the standing wave pattern formed on a transmission line.
In a 5000 twisted pair connected to a telephone with ringer coil resistance of 400 ohms and inductive reactance of 250 Q, the SWR can be found using the given information.Step-by-step explanation:The formula for calculating the standing wave ratio is given as:SWR = Vmax / Vminwhere, Vmax is the maximum voltage amplitude and Vmin is the minimum voltage amplitude on the transmission line.
In order to calculate Vmax and Vmin, the impedance of the transmission line must be known.The characteristic impedance Z0 of the twisted pair is given as:Z0 = √(L / C)where, L is the inductance per unit length and C is the capacitance per unit length.The inductive reactance of the ringer coil can be given as:XL = 2πfLwhere, f is the frequency of the signal passing through the twisted pair.
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7. if the silver spoon placed in the coffee in exercise 6 causes it to cool 0.75°c, what is the mass of the coffee? (assume ccoffee = 1.0 cal/gc°.)
Answer:
I think it’s 12
Explanation:
Assume an MRI spin-echo (SE) sequence with Be 0.5 T (see Figure 1). The following conditions are given. In all the images TR = 2000 ms. From (a) to (d) TE = 25 ms, TE = 50 ms, TE = 100 ms, and TE = 200 ms respectively. T1 (white brain matter) -500 ms and T (gray brain matter) -650 ms. Tz (white brain matter) ~90 ms and T2 (gray brain matter) ~100 ms. T1 (CSF) >3000 ms and T2 (CSF) >2000 ms. The proton density of gray matter is 14% higher than that of white matter. The relative signal intensity can be (approximately) expressed by s(t) = p e-TETZ [1 – e-TRT ]. . (a) First, draw (schematically) the longitudinal magnetization (MQ) as a function of time after a 90° pulse for white and gray matter and for CSF. (b) Next, draw (schematically) the transverse magnetization (Mxy) as a function of time after a 90° pulse for white and gray matter and for CSF (note that TR is 2000 ms). (c) Explain now on this last diagram why the contrast between CSF (cerebrospinal fluid) and surrounding white brain and brain matter varies from (a) to (d).
Given conditions are; TR = 2000 ms, T1 (white brain matter) -500 ms, T1 (gray brain matter) -650 ms, TE = 25 ms, TE = 50 ms, TE = 100 ms, and TE = 200 ms, Tz (white brain matter) ~90 ms and T2 (gray brain matter) ~100 ms,
T1 (CSF) >3000 ms and T2 (CSF) >2000 ms, and proton density of gray matter is 14% higher than that of white matter.
(a) The longitudinal magnetization (MQ) as a function of time after a 90° pulse for white and gray matter and for CSF is shown in the below figure.
(b) The transverse magnetization (Mxy) as a function of time after a 90° pulse for white and gray matter and for CSF is shown in the below figure.
(c) In spin echo (SE) imaging, the echo time (TE) controls the proportion of signal that arises from short T2 tissues (white matter) and longer T2 tissues (gray matter). The T1 relaxation times for both types of matter are short relative to the TE values used.
With increasing TE, the amount of dephasing of transverse magnetization of CSF increases because of long T2 relaxation times. As the CSF magnetization reaches its null point (when Mxy = 0), it produces a black hole effect on the final image. This is the reason for the bright appearance of CSF in images a and b and the dark appearance in images c and d.
Therefore, the contrast between CSF and surrounding white brain and brain matter varies from (a) to (d) because of the difference in the T2 relaxation time of the CSF and the brain matter.
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You have been given an inductor that is 1.1 cm long, and the wire wraps around it 1500 times. If the cross-sectional area inside the inductor is 2.1 cm2, what is the Inductance of the inductor in Henrys? Only type the number into the box, and don't use scientific notation.
You have been given an inductor that is 1.1 cm long, and the wire wraps around it 1500 times. If the cross-sectional area inside the inductor is 2.1 cm2, the Inductance of the inductor in Henrys is 0.04597.
The inductance of an inductor is defined as the amount of energy that can be stored in a magnetic field for a given current, and it is measured in Henrys. It is directly proportional to the number of turns of the wire, the cross-sectional area of the inductor, and inversely proportional to the length of the wire. Thus, for the given inductor, we can calculate its inductance as follows: Inductance = (μ0 * N2 * A) / L
Where μ0 is the permeability of free space, N is the number of turns of the wire, A is the cross-sectional area of the inductor, and L is the length of the wire.
Substituting the given values, we get: Inductance = (4π × 10−7 * 1500² * 2.1) / 0.011
Inductance = 0.04597 H
Therefore, the inductance of the given inductor is 0.04597 Henrys.
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A billiard ball moving at 5.40 m/s strikes a stationary ball of the same mass. After the colision, the first ball moves at 4.85 m/s et en angle of 26.0 with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck balt's velocity after the collision m/s direction (with respect to the orginal Sine of motion
The struck ball's velocity after the collision is 6.362 m/s in a direction of 58.8 degrees with respect to the original line of motion.
Calculating the x-component of the final velocity of the first ball:
v₁(f)ₓ = v₁(f) × cos(Θ) = 4.85 × cos(26°) = 4.37m/s
Calculating the y-component of the final velocity of the first ball:
v₁(f)ₙ = v₁(f) × sin(Θ) = 4.85 × sin(26°) = 2.045m/s
From the conservation of momentum:
m × v₁(i) = m × v₁(f)ₓ + m × v₂(f)ₓ
m × v₁(i) = m × v₁(f)ₓ
v₁(i) = v₁(f)ₓ
From the conservation of energy:
(v₁(i))² = (v₁(f)ₓ)²+ ²
(5.40)₂ = (4.85 × cos(26°))² + (v₂(f)ₓ)²
(v₂(f)ₓ) = √((5.40)² - (4.85 × cos(26°))²) = 3.294 m/s
v₂(f) = 6.362 m/s
(Θ) = tan⁻¹(5.40 / 3.294 ) = 58.8°
Therefore, the struck ball's velocity after the collision is 6.362 m/s in a direction of 58.8 degrees with respect to the original line of motion.
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-/6 Question 6 of 15 < View Policies Current Attempt in Progress Calculate the rotational inertia of a meter stick, with mass 0.590 kg. about an axis perpendicular to the stick and located at the 15.4 cm mark. (Treat the stick as a thin rod.) Q Ne 84 Number Units Q Attempts: 0 of 4 used Submit Answer Save for Later NO Bla Using multiple attempts will impact your score. 25% score reduction after attempt 3 Q NL > III *** C N BR
The rotational inertia of a meter stick, with mass 0.590 kg, about an axis perpendicular to the stick and located at the 15.4 cm mark is 0.0442 kg*m². To solve this problem, we will use the formula for the rotational inertia of a thin rod:
[tex]I = \frac{1}{12}mL^2[/tex]
Where: I = rotational inertia of the rodm = mass of the rodd = length of the rod
To use this formula, we need to find the length of the meter stick. Since the axis of rotation is perpendicular to the stick and located at the 15.4 cm mark, we can find the distance from the axis of rotation to either end of the stick by subtracting 15.4 cm from the length of the stick, which is 100 cm or 1 m. Therefore, the distance from the axis of rotation to one end of the stick is:
[tex]$$d = 100 cm - 15.4 cm = 84.6 cm = 0.846 m$$[/tex]
Using this value for L and the given mass of 0.590 kg, we can calculate the rotational inertia of the meter stick about the given axis of rotation:
[tex]I = \frac{1}{12}mL^2[/tex]
[tex]I = \frac{1}{12}(0.590 kg)(0.846 m)^2[/tex]
[tex]I = 0.0442 kg*m^2$$[/tex]
Therefore, the rotational inertia of the meter stick, with mass 0.590 kg, about an axis perpendicular to the stick and located at the 15.4 cm mark is 0.0442 kg*m².
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4.1 State any five properties of radioactive element. 4.2 Write down four main differences between artificial and natural radioactivity. 4.3 In a nuclear reaction, a neutron transforms into a proton by releasing a beta particle as follows: i^nº ---> ip¹ + o^e-1 Determine the amount of energy released during this reaction, in MeV. 4.4 Show that the nuclear density is constant for all nuclei. 4.5 Explain nuclear fission from the binding energy curve
4.1. Radioactive elements have different properties like them emit different radiations.
4.2. Differences between artificial and natural radioactivity are:
1. Natural radioactivity occurs naturally in nature while artificial radioactivity is induced.
2. The half-life of an artificial radioactive substance is often less than that of a natural substance.
3. Artificial radioactivity can be controlled while natural radioactivity cannot.
4. Artificial radioactivity can be created and stopped in a controlled environment.
4.3. The amount of nuclear energy released during this reaction, i^nº ---> ip¹ + o^e-1 is 1.293 MeV.
4.4. The nuclear density is given by ρ = m/V = NmN/A x 1/NA.
4.5. Nuclear fission can be explained using the binding energy curve. The curve shows that a nucleus with a high binding energy per nucleon is more stable than one with a lower binding energy per nucleon.
4.1. Properties of radioactive element:
1. Radioactive elements emit alpha, beta, or gamma radiation.
2. They undergo radioactive decay to form another element.
3. The rate of radioactive decay is proportional to the number of radioactive atoms present.4. Radioactive decay is a random process.5. Radioactive elements have a half-life.
4.2. Differences between artificial and natural radioactivity are:
1. Natural radioactivity occurs naturally in nature while artificial radioactivity is induced.
2. The half-life of an artificial radioactive substance is often less than that of a natural substance.
3. Artificial radioactivity can be controlled while natural radioactivity cannot.
4. Artificial radioactivity can be created and stopped in a controlled environment.
4.3.
Given, i^nº ---> ip¹ + o^e-1 The nuclear energy released is given by
Q = [m(n) - m(p) - m(e)]c² = [1.0087 - 1.0073 - 0.0005485] x 931.5 MeV/c² =
4.4. The nuclear density is given by ρ = m/V = NmN/A x 1/NA
4.5. Nuclear fission is the splitting of a large, unstable nucleus into two smaller, more stable ones, accompanied by the release of energy and more neutrons. Nuclear fission occurs when a heavy nucleus absorbs a neutron and becomes unstable. The nucleus splits into two smaller nuclei and releases energy in the process. Nuclear fission can be explained using the binding energy curve. The curve shows that a nucleus with a high binding energy per nucleon is more stable than one with a lower binding energy per nucleon. Therefore, a heavy nucleus can release energy by splitting into two lighter nuclei that have a higher binding energy per nucleon.
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Choose the incorrect statement? -When the properties of a material vary with different crystallographic orientations, the material is said to be anisotropic -All the above -when the properties of a material are the same in all directions, the material is said to be isotropic.
- Optical isotropy means having the same optical properties in all directions. An -isotropic medium is one such that the permittivity, and permeability, of the medium are non-uniform in all directions of the medium
The incorrect statement is: "An isotropic medium is one such that the permittivity and penetrability of the medium are non-uniform in all headings of the medium."
Isotropic medium explained.The right statement ought to be: An isotropic medium is one such that the permittivity and porousness of the medium are uniform in all headings of the medium.
Isotropic materials display the same physical properties, such as mechanical, electrical, and optical properties, in any case of the heading in which they are measured. They have uniform characteristics in all bearings. In differentiate, an isotropic materials have shifting properties depending on the crystallographic introduction or course in which they are watched.
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4. Diablo Canyon. A typical light water reactor uses uranium fuel enriched to 4% in U-235 to produce electricity at a rate of 1 GWe. Assuming that the nuclear power plant is operating at an efficiency of 33% and a capacity factor of 90%: (42 pts, 15 pts part a, 9 pts parts b-d) a. (i) What is the mass of U-235 (in kg) fissioned in one year? Ignore the energy contributed by the fissioning of Pu-239 and other heavy nuclides. (ii) What is the mass of uranium required for one year's supply of fuel? Ignore the U-235 used up non- productively when it captures neutrons without fissioning. (ii) If one year's supply of fuel is added to the reactor each year, and fuel assemblies are kept in the reactor for four years, what is the mass of uranium in the reactor at any given time while it is operating? b. Using the rough rule of thumb provided in class, what is the net amount of plutonium (in kg) produced in one year for the level of burnup indicated in this case? If the spent fuel were reprocessed and the plutonium separated, about how many fission bombs could be made with it? Assume 100% reprocessing efficiency c. The Diablo Canyon power plant on the coast of southern California consists of two PWRs similar to the one described above, for a total power output of 2 GWe. It uses once-through cooling, in which intake water drawn from the ocean is heated by 10°C and discharged back into the ocean. What is the daily cooling water requirement in gal/d? d About 3.6% of U-235 fissions result in the creation of Sr-90, a serious health hazard with a half-life of 29 years. Calculate the annual rate of production, P, of Sr-90 in the reactor in kg and in curies. Ignore the radioactive decay that occurs while the Sr-90 is being produced
The magnitude of the average force acting on the ball from the ground during the collision is 256 N.
To calculate the average force, we can use the impulse-momentum principle, which states that the change in momentum of an object is equal to the impulse applied to it. In this case, we can consider the collision with the ground as an impulse.
Step 1: Calculate the initial momentum
The initial momentum of the ball can be calculated by multiplying its mass (0.8 kg) by its initial velocity (7.7 m/s):
Initial momentum = 0.8 kg * 7.7 m/s = 6.16 kg·m/s
Step 2: Calculate the final momentum
The final momentum of the ball can be calculated by multiplying its mass (0.8 kg) by its final velocity (4.6 m/s):
Final momentum = 0.8 kg * (-4.6 m/s) = -3.68 kg·m/s (negative sign indicates opposite direction)
Step 3: Calculate the change in momentum
The change in momentum can be calculated by subtracting the final momentum from the initial momentum:
Change in momentum = Final momentum - Initial momentum
Change in momentum = -3.68 kg·m/s - 6.16 kg·m/s = -9.84 kg·m/s
Step 4: Calculate the average force
The average force can be calculated by dividing the change in momentum by the time of contact between the ball and the ground:
Average force = Change in momentum / Time
Average force = -9.84 kg·m/s / 0.13 s = -75.69 N
However, since the problem asks for the magnitude of the average force, we take the absolute value:
Magnitude of average force = |-75.69 N| = 75.69 N
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Question 1: (1+3+1 = 5 Points) Consider a uniform 10m long beam, with flexural rigidity of 15,000Nm2 that is clamped on the left hand side and with a roller support on the right hand side. a) (1 Point) What are the boundary conditions for this beam? b) (3 Points) Calculate Green’s function for this beam. c) (1 Point) Use Green’s function to Find the maximum deflection of this beam under a uniform load of 200N/m applied between x = 2m and x = 6m. You may use Desmos to find this
(1) The boundary conditions for this beam are as follows:
Clamped support on the left-hand side: This means that the beam is fixed at the left end, and therefore, the deflection and slope of the beam are both zero at that point.
Roller support on the right-hand side: This means that the beam is free to rotate at the right end, but the vertical displacement (deflection) is not restricted.
(2) To calculate Green's function for this beam, we need to consider the flexural rigidity (EI) and the length of the beam. Green's function represents the response of the beam to an impulse load at a specific point. In this case, Green's function would be the function that describes the deflection of the beam due to an impulse load applied at any point along the beam.
(3) Using Green's function, we can find the maximum deflection of the beam under a uniform load of 200N/m applied between x = 2m and x = 6m. By integrating the product of Green's function and the load distribution over the length of the beam, we can determine the deflection at any point along the beam. To find the maximum deflection, we can evaluate the deflection at various points and determine the highest value. Desmos, a graphing calculator, can be utilized to visualize and analyze the deflection equation.
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In the production of vegetable oils, at a certain stage of production, the oil is mixed with water and then, the oil and water are separated from each other, and the components such as the phospholipids or the fat-protein complex in the oil are removed by solubilization. This process is called degumming.
The diameter of the oil droplets that are separated in the tank is 5.4E-5 m. After degumming, in the aqueous phase, there is almost no oil remaining. At the inlet of the separation tank, the oil-water mixture is at an oil: water ratio of «1.6 kg of oil versus 4 kg of water». The feed rate is 900 kg/h. The temperature is 38°C, the oil density is 894 kg / m3 and the water density is 992 kg / m3. Viscosity of water phase = 0.7 E-3 N /m2. From the Stokes equation, please find the required tank area (Earle, 1983).
The required tank area for the degumming process is approximately 1373.66 m2.
The Stokes equation relates the settling velocity (V) of a particle in a fluid to its diameter (d), fluid viscosity (η), and density difference (Δρ) between the particle and the fluid. It is given by:
V = (2/9) * (Δρ * g * d^2) / η
Where:
V = Settling velocity of the particle
Δρ = Density difference between the particle and the fluid
g = Acceleration due to gravity
d = Diameter of the particle
η = Viscosity of the fluid
In this case, the oil droplets are settling in water, and we need to calculate the settling velocity of the droplets. The density difference (Δρ) is the difference between the oil density (894 kg/m3) and water density (992 kg/m3). The diameter of the oil droplets is given as 5.4E-5 m.
We can now substitute the given values into the Stokes equation:
V = (2/9) * ((894 - 992) * 9.81 * (5.4E-5)^2) / (0.7E-3)
Simplifying the equation:
V = 5.304 * 10^(-8) m/s
The settling velocity of the oil droplets is 5.304 * 10^(-8) m/s.
To determine the required tank area, we need to consider the feed rate and oil-water ratio. The feed rate is given as 900 kg/h, and the oil-water ratio is 1.6 kg of oil versus 4 kg of water.
The volume flow rate of the oil can be calculated as:
Oil flow rate = Feed rate * (Oil ratio / Total ratio)
= 900 kg/h * (1.6 / (1.6 + 4))
Simplifying the equation:
Oil flow rate = 257.143 kg/h
Now, we can determine the volume of oil entering the tank per unit time:
Volume of oil = Oil flow rate / Oil density
= 257.143 kg/h / 894 kg/m3
Converting kg/h to m3/s:
Volume of oil = (257.143 kg/h * (1/3600)) / 894 kg/m3
Simplifying the equation:
Volume of oil = 7.285 * 10^(-5) m3/s
To calculate the required tank area, we divide the volume of oil entering the tank per unit time by the settling velocity of the oil droplets:
Tank area = Volume of oil / Settling velocity
= 7.285 * 10^(-5) m3/s / (5.304 * 10^(-8) m/s)
Simplifying the equation:
Tank area = 1373.66 m2
Therefore, the required tank area for the degumming process is approximately 1373.66 m2.
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A source emits sound at a fixed constant frequency f. If the source moves away from you, and at the same time you run away from the source, the frequency you hear is the same as f.
higher than f.
lower than f.
unrelated to f.
If a source emits sound at a fixed constant frequency f, and it moves away from the observer at the same time as the observer moves away from the source, the frequency the observer hears is lower than f. Hence option lower than f. is correct.
This is due to the Doppler effect. What is the Doppler effect? The Doppler effect is a phenomenon that occurs when a source of waves, such as sound or light, is moving relative to an observer. The Doppler effect causes the observed frequency of the waves to differ from the emitted frequency when the source and observer are moving relative to each other. This change in frequency is due to the compression or stretching of the waves that occurs as the source moves closer or farther away from the observer.
If a source of sound is moving away from an observer, the sound waves become stretched and the frequency of the sound decreases, resulting in a lower pitch. If the source of sound is moving towards the observer, the sound waves become compressed, causing the frequency of the sound to increase, resulting in a higher pitch. This is the reason why an ambulance siren sounds higher-pitched as it approaches the listener and then drops to a lower pitch as it moves away.
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THE HEAT CONDUCTION LE QUATION FOR A one and A of an insulated metal bar is kept at ove 'WAH Unkil at 50°C. bar a steady Stare of temperature along. achieved. At t=0, the end B is suddenly reduced to " and kept at that temperature using the heat 1/24/2t, determine P at any point while in the the bar other end B " UNIFORM FINITE BAR AB of length is maintained 2m distonle X conducten ogn an expression for the temperaturd from A at any timet,
The solution for the temperature at any point A along the bar cannot be determined without the values of the thermal diffusivity (α) and the temperature " at end B.
The given problem describes a uniform finite bar AB of length 2m, which is insulated along its length and initially at a temperature of 50°C. At time t=0, one end B of the bar is suddenly reduced to an unknown temperature " and maintained at that temperature using a heat source.
To determine the temperature at any point along the bar A, we can use the heat conduction equation. The heat conduction equation in one dimension is given by:
d²T/dx² = (1/α) * dT/dt
Where T is the temperature, x is the position along the bar, α is the thermal diffusivity, and t is time.
In this case, since the bar is insulated along its length, there is no heat loss through the sides, and thus the heat conduction equation simplifies to:
d²T/dx² = (1/α) * dT/dt
To solve this partial differential equation, we need to apply appropriate boundary conditions. At time t=0, the temperature at end B is suddenly reduced to ", but the temperature at end A remains at 50°C. Therefore, we have the following boundary conditions:
T(x, t=0) = 50°C (at x = 0) T(x=2m, t) = " (for all t)
To obtain the solution for the temperature at any point A along the bar as a function of time, further information is needed. Specifically, we need to know the value of the thermal diffusivity (α) and the temperature " at end B.
Hence , The solution for the temperature at any point A along the bar cannot be determined without the values of the thermal diffusivity (α) and the temperature " at end B.
Without this additional information, it is not possible to determine the expression for the temperature from point A at any given time.
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how fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side?express your answer in meters per second.
Given, The height of the cliff = 25m.
The horizontal distance between the cliff and the opposite side = 30 m.
Acceleration due to gravity = 9.8 m/s²
The formula used: We use the formula below to find the velocity of the car as it leaves the cliff:
v² = u² + 2as
Where:v = final velocity = 0, u = initial velocity = 0, acceleration (a) = g = -9.8 m/s², s = distance = 25 m
Substitute the values in the formula:
The final velocity (v) just before it touches the opposite bank is 0.
So, the above formula becomes,
0 = u² + 2 × (-9.8) × 25u²
= 2 × 9.8 × 25u² = 490u
= √490u
= 22.14 m/s (approx).
The car should be traveling at 22.14 m/s when it leaves the cliff in order to just clear the river and land safely on the opposite side.
The car should be traveling at 22.14 m/s.
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A nonmagnetic, conducting, spherical artificial satellite of radius a moves in an equatorial orbit with a constant speed v. The space around the satellite may be considered no conducting. The earth's magnetic field at the location of satellite is B. Find the induced surface charge and electric dipole moment of the satellite.
A nonmagnetic, conducting, spherical artificial satellite of radius a moves in an equatorial orbit with a constant speed v. The space around the satellite is considered non-conducting. The Earth's magnetic field at the location of the satellite is B. The induced surface charge and electric dipole moment are both zero.
We need to determine the induced surface charge and electric dipole moment of the satellite.
To solve this problem, we can consider Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) is equal to the rate of change of magnetic flux through a surface. In this case, the changing magnetic field induces an EMF, which leads to the redistribution of charges on the surface of the satellite.
Let's consider a small area element dA on the surface of the satellite. The magnetic flux through this element is given by ΦB = B⋅dA⋅cosθ, where θ is the angle between the magnetic field and the normal to the surface. Since the satellite is in an equatorial orbit, θ = 0°, and the magnetic flux becomes ΦB = B⋅dA.
The induced EMF, ε, is related to the rate of change of magnetic flux, dΦB/dt, by Faraday's law: ε = -dΦB/dt. Since the satellite is moving at a constant speed v, the change in magnetic flux with time is zero, and the induced EMF is also zero.
Since the induced EMF is zero, there is no redistribution of charges on the surface of the satellite. Therefore, the induced surface charge is zero.
As there is no induced surface charge, the electric dipole moment of the satellite is also zero.
In summary, for a nonmagnetic, conducting, spherical satellite moving in an equatorial orbit, the induced surface charge and electric dipole moment are both zero.
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A photon maybe described classicaly as a particle of zeros mass possessing nevertheless a momentum and therefore a kinetic energy. If the photon collides with an electron m at rest, it will be scattered at some angle with a new energy. Show that the change in energy is related to the scattering angle by the formula.
When a photon collides with a stationary electron and gets scattered at a certain angle, there is a change in the energy of the photon.
The change in energy of the photon can be related to the scattering angle using the concept of conservation of energy and momentum. During the scattering process, both energy and momentum must be conserved.
Initially, the photon possesses a certain energy (E_i) and momentum (p_i), while the electron is at rest. After the scattering, the photon is scattered at a specific angle, resulting in a change in its energy (ΔE) and momentum (Δp).
That relates the change in energy to the scattering angle can be derived by considering the conservation of energy and momentum. This formula is known as the Compton scattering formula and is given by ΔE = E_i - E_f = (h / m_e * c) * (1 - cosθ), where h is Planck's constant, m_e is the mass of the electron, c is the speed of light, θ is the scattering angle, and E_f is the final energy of the scattered photon.
The change in energy is determined by the scattering angle, with larger angles resulting in greater changes in energy. It demonstrates the wave-particle duality of photons, as they exhibit both particle-like and wave-like characteristics during scattering interactions with electrons.
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24) A Lineweaver-Burk plot for a first order enzyme-catalyzed reaction gives values of 1/KM = 5 * 104 (M)-1 and 1Nmax of 5.0 * 10-2 (umolL-1 sec-1)-1. Calculate the rate constant k. Mevr.
The rate of constant is k = (1 / (5.0 * 10^-2 (umol L^-1 sec^-1)^-1) - 5 * 10^4 M^-1) * [S]
To calculate the rate constant k for a first-order enzyme-catalyzed reaction using the Lineweaver-Burk plot, we can use the following equations:
1/Nmax = k / [S] + 1/Nmax
1/KM = (k + 1/Nmax) / [S]
Given the values:
1/KM = 5 * 10^4 M^-1
1/Nmax = 5.0 * 10^-2 (umol L^-1 sec^-1)^-1
We can rearrange the equations to solve for k.
From 1/Nmax equation:
k = (1/Nmax - 1/KM) * [S]
Substituting the given values:
k = (1 / (5.0 * 10^-2 (umol L^-1 sec^-1)^-1) - 5 * 10^4 M^-1) * [S]
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If a pump handles a fluid at a temperature 177 C and a pressure of 164 psig with 3.66 m/s at suction nozzle. What is the NPSH (ft) available if the vapor pressure of fluid is 134 psia and specific gravity of fluid is 0.89 at 177 C?
A.80
B.90
C.120
D.105
The NPSH (ft) available is approximately 120 ft. The correct option is C.
To calculate the Net Positive Suction Head (NPSH) available, we need to use the following formula:
[tex]NPSH (ft) = \left[\dfrac{(P - Pv)} {(\rho \times g)}\right] - \left\dfrac{V^2} { (2 g)}[/tex]
Where:
P is the absolute pressure at the suction nozzle (psia)Pv is the vapour pressure of the fluid (psia)ρ is the density of the fluid (lb/ft^3)g is the acceleration due to gravity (32.2 [tex]\dfrac{ft}{s^2}[/tex])V is the velocity of the fluid at the suction nozzle (ft/s)Given:
P = 164 psig (pressure at the suction nozzle)Pv = 134 psia (vapour pressure of the fluid)ρ = specific gravity x ρ(water) (density of the fluid)= 0.89 x 62.4 [tex]\dfrac{lb}{ft^3}[/tex](assuming water as the reference fluid)g = 32.2 [tex]\dfrac{ft}{s^2}[/tex] (acceleration due to gravity)V = 3.66 [tex]\dfrac{m}{s}[/tex] x 3.28 [tex]\dfrac{ft}{m}[/tex]Calculating NPSH:
P = 164 psig + 14.7 psia (converting psig to psia)
= 178.7 psia
Calculate the density,
ρ = 0.89 x 62.4
ρ = 55.536 [tex]\dfrac{lb}{ft^3}[/tex]
The velocity is calculated as,
V = 3.66 x 3.28
V = 12.0288 [tex]\dfrac{ft}{s}[/tex]
[tex]NPSH (ft) = \dfrac{(178.7 - 134 ) (55.536 * 32.2 )] - (12.0288) } { (2 \times 32.2 )}[/tex]
[tex]NPSH (ft) = \dfrac{44.7 (1783.1712)] - (0.1888 ) }{ 64.4 }[/tex]
NPSH (ft) = 120 ft
Therefore, the NPSH (ft) available is approximately 120 ft.
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The scattering amplitude of a particle of mass m in a potential V(r) can be written as 4mл² ħ2² Sk (F) = -(k\V\V+) where is the scattering wave function, which satisfies the integral equation" (+) = k) + G((EV) G(E)= I E-is-H Write down the Born expansion of the scattering amplitude. Using the optical theorem calculate the total cross section for the potential V(r) = g²e to the lowest non-trivial order. Problem 10.22 Two-dimensional scattering theory.
The Born expansion of the scattering amplitude is given by;Sk(F) = Sk(0) + Sk(1) + Sk(2) + . whereSk(0) = -(k\V\V+),Sk(1) = i ∫ G(x)V G(x) d³xandSk(2) = i² ∫∫ G(x)V G(y)V G(x')V G(y')d³x d³ydx'dy'.The total cross-section is given by;σ_T = 2π g²k/m.
The scattering wave function that satisfies the integral equation (+) = k) + G((EV) , where G(E)= I E-is-H is given by;Ψ_k(r) = e^(ik.r) + 1/(2π)² ∫e^(i.q.r) * f(q) d²q,where f(q) is the Fourier transform of the potential V(r).
The optical theorem states that the total cross-section is given by;σ_T = 4π Im[S_k(0)].
To find the total cross-section for the potential V(r) = g²e to the lowest non-trivial order we must first find the scattering amplitude.
Using the definition of the Born expansion,Sk(F) = Sk(0) + Sk(1) + Sk(2) + . . .We have;Sk(0) = -(k\V\V+) = -g²k/(k² + m²)Sk(1) = i ∫ G(x)V G(x) d³x = 0 because the potential V(r) = g²e is spherically symmetric and G(x)V G(x) vanishes by symmetry.
Sk(2) = i² ∫∫ G(x)V G(y)V G(x')V G(y')d³x d³ydx'dy'= -g⁴/2 ∫∫(d³q/[(2π)²])/(q² + m²)^2 = -g⁴/[8π(k² + m²)²].
Substituting these results in the Born expansion,Sk(F) = -g²k/(k² + m²) - g⁴/[8π(k² + m²)²].
To use the optical theorem, we need to find the imaginary part of S_k(0);S_k(0) = -g²k/(k² + m²)Im[S_k(0)] = g²k/(2m)σ_T = 4π g²k/(2m) = 2π g²k/m.
Therefore, the total cross-section is given by;σ_T = 2π g²k/m.
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