The net gravitational force on the third mass, located above the first two masses in an equilateral triangle formation, is zero. This means that the gravitational forces exerted by the first two masses cancel each other out.
The gravitational force between two masses can be calculated using Newton's law of universal gravitation: F = G * (m1 * m2) / r², where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses, and r is the distance between the masses.
In this case, the first and second masses are located at the origin and x = 33 cm, respectively. Since the masses are identical and the triangle formed is equilateral, the distance between the first and second masses is also 33 cm.
The gravitational force between the first and second masses is given by F1 = G * (m * m) / (0.33)^2, and it acts along the line joining these masses. Since the triangle is equilateral, the third mass is located directly above the midpoint between the first two masses.
As a result, the gravitational force exerted by the first mass on the third mass is equal in magnitude but opposite in direction to the gravitational force exerted by the second mass on the third mass. Therefore, these two forces cancel each other out, resulting in a net gravitational force of zero on the third mass.
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An object's velocity follows the equation = 3+2 +1. What is the object's displacement as a function of time?
The object's displacement as a function of time can be found by integrating its velocity equation with respect to time.The object's displacement as a function of time is x(t) = t^3 + t^2 + t + C.
The velocity equation is given as v(t) = 3t^2 + 2t + 1. To find the object's displacement, we integrate this equation with respect to time.Integrating v(t) gives us the displacement equation x(t) = ∫(3t^2 + 2t + 1) dt. Integrating term by term, we get x(t) = t^3 + t^2 + t + C, where C is the constant of integration.
Therefore, the object's displacement as a function of time is x(t) = t^3 + t^2 + t + C. By integrating the given velocity equation with respect to time, we find the displacement equation. Integration allows us to find the antiderivative of the velocity function, which represents the change in position of the object over time.
The constant of integration (C) arises because indefinite integration introduces a constant term that accounts for the initial condition or starting point of the object.
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A 190,000 kg space probe is landing on an alien planet with a gravitational acceleration of 5.00. If its fuel is ejected from the rocket motor at 40,000 m/s what must the mass rate of change of the space ship (delta m)/( delta t ) be to achieve at upward acceleration of 2.50 m/s ^ 2 ?
A roller coaster cart of mass 114.0 kg is pushed against a launcher spring with spring constant 550.0 N/m compressing it by 11.0 m in the process. When the roller coaster is released from rest the spring pushes it along the track (assume no friction in cart bearings or axles and no rolling friction between wheels and rail). The roller coaster then encounters a series of curved inclines and declines and eventually comes to a horizontal section where it has a velocity 7.0 m/s. How far above or below (vertical displacement) the starting level is this second (flat) level? If lower include a negative sign with the magnitude.
The mass rate of change of the space ship is 190,000 kg/s and the required displacement is 8.88 m (upwards).
Question 1A The space probe lands on an alien planet with a gravitational acceleration of 5.00 m/s².
Now, the upward acceleration required is 2.50 m/s². Hence, the required acceleration can be calculated as:
∆v/∆t = a Where,
∆v = change in velocity = 40,000 m/s
a = acceleration = 2.50 m/s²
∆t can be calculated as:
∆t = ∆v/a
= 40,000/2.5
= 16,000 seconds
Therefore, the mass rate of change of the space ship is calculated as:
∆m/∆t = (F/a)
Where, F = force
= m × a
F = (190,000 kg) × (2.5 m/s²)
F = 475,000 N
∆m/∆t = (F/a)∆m/∆t
= (475,000 N) / (2.5 m/s²)
∆m/∆t = 190,000 kg/s
Question 2 Mass of the roller coaster, m = 114 kg
Spring constant, k = 550 N/m
Compression, x = 11.0 m
Initial velocity of the roller coaster, u = 0
Final velocity of the roller coaster, v = 7.0 m/s
At point A (Start)
Potential Energy + Kinetic Energy = Total Energy
[tex]1/2 kx^2+ 0 = 1/2 mv^2 + mgh[/tex]
[tex]0 + 0 = 1/2 \times 114 \times 7^2 + 114 \times g \times h[/tex]
[tex]1/2 \times 114 \times 7^2 + 0 = 114 \times 9.8 \times h[/tex]
h = 16.43 m
At point B (End)
Potential Energy + Kinetic Energy = Total Energy
[tex]0 + 1/2 \ mv^2 = 1/2 \ mv^2 + mgh[/tex]
[tex]0 + 1/2 \times 114 \times 7^2= 0 + 114 \times 9.8 \times h[/tex]
h = -7.55 m
So, the vertical displacement is 16.43 m - 7.55 m = 8.88 m (upwards)
Therefore, the required displacement is 8.88 m (upwards).
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The vertical displacement from the starting level to the second (flat) level.
To determine the mass rate of change of the space ship (Δm/Δt) needed to achieve an upward acceleration of 2.50 m/s², we can use the rocket equation, which states:
Δv = (ve * ln(m0 / mf))
Where:
Δv is the desired change in velocity (2.50 m/s² in the upward direction),
ve is the exhaust velocity of the fuel (40,000 m/s),
m0 is the initial mass of the space probe (190,000 kg + fuel mass),
mf is the final mass of the space probe (190,000 kg).
Rearranging the equation, we get:
Δm = m0 - mf = m0 * (1 - e^(Δv / ve))
To find the mass rate of change, we divide Δm by the time it takes to achieve the desired acceleration:
(Δm / Δt) = (m0 * (1 - e^(Δv / ve))) / t
To determine the vertical displacement of the roller coaster from its starting level when it reaches the second (flat) level with a velocity of 7.0 m/s, we can use the conservation of mechanical energy. At the starting level, the only form of energy is the potential energy stored in the compressed spring, which is then converted into kinetic energy at the second level.
Potential energy at the starting level = Kinetic energy at the second level
0.5 * k * x^2 = 0.5 * m * v^2
where:
k is the spring constant (550.0 N/m),
x is the compression of the spring (11.0 m),
m is the mass of the roller coaster cart (114.0 kg),
v is the velocity at the second level (7.0 m/s).
Plugging in the values:
0.5 * (550.0 N/m) * (11.0 m)^2 = 0.5 * (114.0 kg) * (7.0 m/s)^2
Solving this equation will give us the vertical displacement from the starting level to the second (flat) level.
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6. The electric potential function in a volume of space is given by V(x,y,z) = x2 + xy2 + 2yz?. Determine the electric field in this region at the coordinate (3,4,5).
The electric potential function in a volume of space is given by V(x,y,z) = x2 + xy2 + 2yz.The electric field in the region at the coordinate (3, 4, 5) is -22i - 24j - 8k.
To determine the electric field in the given region, we need to calculate the gradient of the electric potential function V(x, y, z) at the coordinate (3, 4, 5).The gradient of a scalar function is a vector that points in the direction of the steepest increase of the function and its magnitude represents the rate of change of the function in that direction.
The electric potential function is given as V(x, y, z) = x^2 + xy^2 + 2yz.
To find the gradient, we need to calculate the partial derivatives of V with respect to each coordinate (x, y, z):
∂V/∂x = 2x + y^2
∂V/∂y = 2xy
∂V/∂z = 2y
Now, we can evaluate these partial derivatives at the coordinate (3, 4, 5):
∂V/∂x = 2(3) + (4)^2 = 6 + 16 = 22
∂V/∂y = 2(3)(4) = 24
∂V/∂z = 2(4) = 8
Therefore, the electric field at the coordinate (3, 4, 5) is given by the vector E = -(∂V/∂x)i - (∂V/∂y)j - (∂V/∂z)k:
E = -22i - 24j - 8k
So, the electric field in the region at the coordinate (3, 4, 5) is -22i - 24j - 8k.
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Two forces, each of magnitude P, are applied to the wrench. The diameter of the steel shaft AB is 30 mm. Determine the largest allowable value of P if the shear stress in the shaft is not to exceed 120 MPa and its angle of twist is limited to 7 deg. Use G=83 GPa for steel B F 600 mm -300 mm
Shear stress is the quotient of a shearing force by the area parallel to it, defined as force per unit area acting parallel to the plane .The angle of twist is the degree of deformation that occurs as a result of twisting forces on a body. The maximum allowable value of P is 102.9 N.
When an external torque or moment is applied to a shaft, it produces shear stresses and angles of twist. Now, let us consider the given scenario. The magnitude of two forces P is applied to a wrench, and the diameter of the steel shaft AB is 30 mm. To determine the largest allowable value of P, we must first calculate the maximum shear stress and the angle of twist .Because shear stress is calculated as
τ = P/(π/4) x d², we can rearrange it to find P, which is P = τ x (π/4) x d².The largest allowable value of P can be determined if the shear stress is limited to 120 MPa and the angle of twist is limited to 7 degrees.
Maximum shear stress can be calculated using τmax = (16/3) x T / π x d³, where T is the applied torque. The angle of twist is calculated as Δθ = TL/GJ, where TL is the total torque and J is the polar moment of inertia.
Considering the formulae mentioned above, we have;
τmax = (16/3) x T / π x d³120 x 10⁶ = (16/3) x T / π x (30 x 10⁻³)³
=> T = 3147.4
NmΔθ = TL/GJ7 x (π/180) = (3147.4 x 0.6) / (83 x 10⁹ x π/32 x (0.3⁴ - 0.28⁴))
=> Δθ = 0.0055 rad
Now, let us calculate P:P = τ x (π/4) x d² => P = 120 x 10⁶ x (π/4) x (30 x 10⁻³)²P = 102.9 N
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13. The photoelectric effect is (a) due to the quantum property of light (b) due to the classical theory of light (c) independent of reflecting material (d) due to protons. 14. In quantum theory (a) t
13. The photoelectric effect is (a) due to the quantum property of light.
The photoelectric effect is a phenomenon in which electrons are emitted from a material when it is exposed to light. This effect can only be explained by considering light as consisting of discrete packets of energy called photons, which is a fundamental concept in quantum theory.
According to the quantum property of light, each photon carries a specific amount of energy, and when it interacts with matter, it can transfer this energy to electrons, causing them to be ejected from the material. Therefore, the photoelectric effect is due to the quantum property of light.
14. In quantum theory (a) the wave-particle duality of matter and energy is explained.
In quantum theory, the wave-particle duality of matter and energy is explained. This principle suggests that particles, such as electrons, can exhibit both wave-like and particle-like properties depending on the context of observation.
This duality is a fundamental concept in quantum mechanics, which describes the behavior of particles and energy at the microscopic level. It means that particles can display wave-like characteristics, such as interference and diffraction, as well as particle-like characteristics, such as position and momentum. This concept is central to understanding the behavior of subatomic particles and the interactions between matter and energy.
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A light beam coming from an underwater spotlight exits the water at an angle of 64.8 to the vertical. Y Part A At what angle of incidence does it hit the air-water interface from below the surface? Η ΑΣΦ ? Submit Request Answer Provide Feedback
When the light beam from the underwater spotlight exits the water at an angle of 64.8 degrees to the vertical, it hits the air-water interface from below the surface with an angle of incidence of 25.2 degrees.
The problem involves a light beam coming from an underwater spotlight and exiting the water at an angle of 64.8 degrees to the vertical. We need to determine the angle of incidence at which the light beam hits the air-water interface from below the surface.
By applying the laws of reflection and refraction, we can calculate the angle of incidence. In this case, the angle of incidence is found to be 25.2 degrees.
When light passes from one medium to another, such as from water to air, it undergoes both reflection and refraction. The angle of incidence (θ₁) is the angle between the incident ray and the normal to the interface, and the angle of refraction (θ₂) is the angle between the refracted ray and the normal.
In this problem, the light beam exits the water at an angle of 64.8 degrees to the vertical. The vertical direction is perpendicular to the surface of the water. Therefore, the angle of incidence is given by:
θ₁ = 90° - 64.8° = 25.2°
This means that the light beam, upon hitting the air-water interface from below the surface, makes an angle of incidence of 25.2 degrees with the normal to the interface.
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Using separation of variables method, solve Schrodinger Eq. to find o as a function of time t.
A function or collection of functions will be the solution to a differential equation, which is made up of a function and one or more of its derivatives.
Thus, These equations can be used to represent movement, growth, oscillations, waves, and any other phenomenon with a rate of change.
In some differential equations, the variables must be separated since there may be multiple variables at play and a solution may exist for one or more of them. In a different example, the (y) needs to be isolated on one side of the equation if there are two variables in the equation.
It is necessary to move the second variable (x) to the opposing side of the equation.
Thus, A function or collection of functions will be the solution to a differential equation, which is made up of a function and one or more of its derivatives.
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The RC circuit of has R=7.2kΩ and C=4.0μF. The capacitor is at voltage V0 at t=0, when the switch is closed. Part A
The solution we get is V = 10V * (1 - e-0.01s/29.4μs) = 2.93V.
The step-by-step answer for Part A of the RC circuit problem:
The time constant of the circuit is τ = RC = 7.2kΩ * 4.0μF = 29.4μs.
The voltage across the capacitor at time t = 0.01s is given by the equation
V = V0(1 - e-t/τ) = 10V * (1 - e-0.01s/29.4μs) = 2.93V.
Therefore, the voltage across the capacitor at time t = 0.01s is 2.93V.
Here is a more detailed explanation of each step:
The time constant of an RC circuit is the time it takes for the voltage across the capacitor to reach 63.2% of its final value. The time constant is calculated by multiplying the resistance of the circuit by the capacitance of the circuit.
The voltage across the capacitor at time t is given by the equation V = V0(1 - e-t/τ), where V0 is the initial voltage across the capacitor, t is the time in seconds, and τ is the time constant of the circuit.
In this problem, V0 = 10V, t = 0.01s, and τ = 29.4μs. Substituting these values into the equation, we get V = 10V * (1 - e-0.01s/29.4μs) = 2.93V.
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"A ray of light strikes a surface at ninety degrees, that is, it
is parallel to the normal. The angle of refraction is
A. one hundred and eighty degrees, 180°
B. ninety degrees, 90°.
C. forty-five degrees
When a ray of light strikes a surface at a 90-degree angle, which means it is parallel to the normal, the angle of refraction is 90 degrees (Option B).
When light passes from one medium to another, it usually undergoes refraction, which is the bending of light due to the change in its speed. The angle of refraction is determined by Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the speeds of light in the two media.
However, when a ray of light strikes a surface at a ninety-degree angle, it is parallel to the normal of the surface. In this case, the light does not change its direction upon entering the new medium, and no refraction occurs. The angle of refraction is undefined because there is no bending or change in the direction of the light ray.
Option A (180 degrees) is incorrect because an angle of 180 degrees would mean that the refracted ray is opposite in direction to the incident ray, which is not possible in this case. Option C (45 degrees) is also incorrect because it does not apply to the scenario described, where the incident ray is parallel to the normal.
When a ray of light strikes a surface at a 90-degree angle, the angle of refraction is also 90 degrees. This occurs because the incident ray, being parallel to the normal, does not undergo any change in direction as it passes from one medium to another.
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Answer the question with a cross in the box you think is correct. If you change your mind about an answer, put a line through the box and then mark your new answer with a cross When a guitar string is plucked, a sound of constant frequency is heard. The wave produced on the vibrating guitar string is A. longitudinal and progressive. B. longitudinal and stationary C. transverse and progressive. D. transverse and stationary
The wave produced on the vibrating guitar string is transverse and progressive.
When a guitar string is plucked, it produces a wave that travels along the string. This wave is transverse in nature, meaning that the particles of the medium (the string) vibrate perpendicular to the direction of wave propagation. As the string oscillates up and down, it creates peaks and troughs in the wave pattern, forming a characteristic waveform.
The wave is also progressive, which means it propagates through space. As the plucked string vibrates, the disturbance travels along the length of the string, carrying the energy of the wave with it. This progressive motion allows the sound wave to reach our ears, where we perceive it as a sound of constant frequency.
In summary, when a guitar string is plucked, it generates a transverse and progressive wave. The transverse nature of the wave arises from the perpendicular vibrations of the string's particles, while its progressiveness refers to the propagation of the wave through space, enabling us to hear a sound of constant frequency.
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6. Design an absorption packed tower that is used to reduce NH; in air from a concentration of 0.10 kg'm' to a concentration of 0.0005 kg/ml Given: Column diameter - 3.00 m Operating temperature - 20.0°C Air density at 20.0°C -1.205 kg/m Operating pressure 101.325 kPa For 15 kg NH, per 100 kg H:0 measured partial pressure of NH3 – 15.199 kPa Q-01-10.0 kg's H-0.438 m H=0.250 m Incoming liquid is water free of NH3(x2) = 0 GMW of NH) - 17.030 GMW of Air - 28.970 GMW of H20 - 18.015 Determine 6.1 Mole fraction of pollutant in the gas phase at inlet of tower (y) 6.2 Equilibrium mole fraction of pollutant in the liquid phase (3) 6.3 Slope of equilibrium curve (m) 6.4 Absorption factor (A) 6.5 Height of an absorption packed tower (2)
The objective is to design an absorption packed tower to reduce NH3 concentration in air, and the parameters to be determined are the mole fraction of pollutant in the gas phase at the tower inlet (y), the equilibrium mole fraction of pollutant in the liquid phase (x), the slope of the equilibrium curve (m), the absorption factor (A), and the height of the tower.
What is the objective of the given problem and what parameters are to be determined for designing the absorption packed tower for NH3 removal?The given problem involves the design of an absorption packed tower for removing NH3 from air. The tower should reduce the NH3 concentration from 0.10 kg/m3 to 0.0005 kg/m3.
The operating conditions include a column diameter of 3.00 m, operating temperature of 20.0°C, air density at 20.0°C of 1.205 kg/m3, and operating pressure of 101.325 kPa. The relevant data includes the measured partial pressure of NH3, the flow rate of H2O, and the molecular weights of NH3, air, and H2O.
The objectives are to determine the mole fraction of the pollutant in the gas phase at the inlet of the tower (y), the equilibrium mole fraction of the pollutant in the liquid phase (x), the slope of the equilibrium curve (m), the absorption factor (A), and the height of the absorption packed tower.
These parameters will help in designing an effective tower for NH3 removal.
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Resistor in circuit is made of a length of 14awg iron wire. When
10 V is applied across the resistor wire of length 100m,
what is the reading on the ammeter? The thickness
of 14awg wire is 1.628mm.
The reading on the ammeter would be approximately 2.14 Amperes.
To calculate the reading on the ammeter, we need to determine the resistance of the 14 AWG iron wire. The resistance can be calculated using the formula
[tex]R = ρ * (L / A)[/tex]
where:
R is the resistance,
ρ is the resistivity of the material (in this case, iron),
L is the length of the wire, and
A is the cross-sectional area of the wire.
First, let's calculate the cross-sectional area of the 14 AWG wire. The diameter of the wire can be obtained from the wire gauge size. For 14 AWG, the diameter is approximately 1.628 mm.
The radius (r) can be calculated by dividing the diameter by 2:
r = 1.628 mm / 2 = 0.814 mm = 0.000814 m
The cross-sectional area (A) can be calculated using the formula:
[tex]R = ρ * (L / A)[/tex]
[tex]A = 3.14159 * (0.000814 m)^2 ≈ 2.07678 × 10^(-6) m^2[/tex]
Next, we need to find the resistivity of iron. The resistivity of iron (ρ) is approximately 9.71 × 10^(-8) Ω·m.
Now, we can calculate the resistance (R) using the formula mentioned earlier:
[tex]R = (9.71 × 10^(-8) Ω·m) * (100 m / 2.07678 × 10^(-6) m^2)[/tex]
[tex]R ≈ 4.675 Ω[/tex]
Therefore, with a 10 V potential difference across the 14 AWG iron wire resistor, the reading on the ammeter would be:
[tex]I = V / R[/tex]
[tex]I = 10 V / 4.675 Ω[/tex]
[tex]I ≈ 2.14 A[/tex]
So, the reading on the ammeter would be approximately 2.14 Amperes.
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Consider two strings tuned to the note A (440 Hz), mounted on guitars of the same size. The high-tension string has a diameter of 0.432 mm, and the low-tension string has a diameter of 0.381 mm. The strings are made of the same material, so they have the same density p. The strings can be thought of as long cylinders. What is the ratio of the high tension to the low tension?
The ratio of high tension to low tension is `1.22`.Hence, option D is correct.
Given data: Frequency of both the string,
`f = 440 Hz`
Diameter of high tension string, `d_1 = 0.432 mm
`Diameter of low tension string, `d_2 = 0.381 mm`
The density of both strings is the same.
Let the tension in high tension string and low tension string be `T_1` and `T_2` respectively.
Ratio of tension in both strings:
`T_1/T_2= [(π/4)d_1²p(f₁)²]/[(π/4)d_2²p(f₂)²]`
Here, `f₁ = f₂ = f =
440 Hz`.
So,
`T_1/T_2 = d_1²/d_2² = (0.432)²/(0.381)²
≈ 1.22`
Therefore, the ratio of high tension to low tension is `1.22`.
Hence, option D is correct.
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The ratio of the high-tension to the low-tension is 1.3616:1.Given Data: Diameter of high tension string: d₁ = 0.432 mm Diameter of low tension string:
d₂ = 0.381 mm
The strings are made of the same material, so they have the same density p.
Frequency of both the strings: f = 440 Hz Formula Used:
The tension (T) in a string is given by, T = μf²d²π² Where, μ is the linear density of the string (mass per unit length)d is the diameter of the string f is the frequency of vibration of the stringπ = 3.14 Calculation:
Let the tension in the high-tension string be T₁ and the tension in the low-tension string be T₂ We know that,μ = pA where, p is the density of the string
A = πd²/4 is the cross-sectional area of the string As the strings are made of the same material, they have the same density.
Therefore,μ₁ = μ₂
⇒ pA₁ = pA₂
⇒ A₁ = A₂d₁²
= d₂²
= (0.432 mm)²
= 0.186624 mm²
= A₁A₂
= (0.381 mm)²
= 0.144961 mm²
Therefore, A₁/A₂ = (0.432 mm)²/(0.381 mm)²
= 1.3616/1T₁ = μf²d₁²π²and,T₂ = μf²d₂²π²Dividing these two equations,
T₁/T₂ = μ₁f²d₁²π²/μ₂f²d₂²π²
⇒ T₁/T₂ = d₁²/d₂²
⇒ T₁/T₂ = (0.432 mm)²/(0.381 mm)²
⇒ T₁/T₂ = 1.3616/1
⇒ T₁/T₂ = 1.3616:1
Therefore, the ratio of the high-tension to the low-tension is 1.3616:1.
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If the velocity of sound in a solid is of the order 103 m/s, compare the frequency of the sound wave λ = 20 Å for (a) a monoatomic system and (b) acoustic waves and optical waves in a diatomic system containing two identical atoms (M=m) per unit cell of interatomic spacing 2.2 Å.
For the given conditions, the frequency of the sound wave in optical waves is 4.3 × 1013 Hz.
Given that the velocity of sound in a solid is of the order 103 m/s, and the frequency of the sound wave is λ = 20 Å.
We have to compare the frequency of the sound wave for (a) a monoatomic system and (b) acoustic waves and optical waves in a diatomic system containing two identical atoms (M=m) per unit cell of interatomic spacing 2.2 Å.
(a) Monoatomic system
The relation between the frequency, wavelength, and velocity of sound wave in a solid is given by:
f = v / λ
Where,
f is frequency,
λ is wavelength, and
v is velocity of sound.
The frequency of the sound wave in monoatomic system is
f = 103 / 20 × 10^-10f = 5 × 10^12 Hz
(b) Diatomic system
The diatomic system contains two identical atoms (M=m) per unit cell of interatomic spacing 2.2 Å.
For diatomic system, there are two modes of vibration in a solid:
Acoustic mode and Optical mode.
Acoustic mode
For acoustic waves in a diatomic system, the angular frequency of the wave is given by:
ω = 2Vs × √(sin²(πn/Na)+(1 - sin²(πn/Na)) / 4) / a
Where,
ω is the angular frequency,
Vs is the velocity of sound in a solid,
n is the mode of vibration,
a is the interatomic spacing, and
Na is the number of atoms per unit cell of a crystal.
The frequency of the sound wave in acoustic mode is
f = ω / 2π
The frequency of the sound wave in acoustic mode for diatomic system is
f = Vs × √(sin²(πn/Na)+(1 - sin²(πn/Na)) / 4) / a × (1 / 2π)f
= 103 × √(sin²(πn/2)+(1 - sin²(πn/2)) / 4) / 2.2 × (1 / 2π)
For n = 1, the frequency of the sound wave in acoustic mode is
f = 0.73 × 10^13 Hz
For n = 2, the frequency of the sound wave in acoustic mode is
f = 1.6 × 10^13 Hz
For n = 3, the frequency of the sound wave in acoustic mode is
f = 2.5 × 10^13 Hz
For n = 4, the frequency of the sound wave in acoustic mode is
f = 3.3 × 10^13 Hz
Optical mode
For optical waves in a diatomic system, the angular frequency of the wave is given by:
ω = 2Vs × √(sin²(πn/Na)-(1 - sin²(πn/Na)) / 4) / a
Where,
ω is the angular frequency,
Vs is the velocity of sound in a solid,
n is the mode of vibration,
a is the interatomic spacing, and
Na is the number of atoms per unit cell of a crystal.
The frequency of the sound wave in optical mode is
f = ω / 2π
The frequency of the sound wave in optical mode for diatomic system is
f = Vs × √(sin²(πn/Na)-(1 - sin²(πn/Na)) / 4) / a × (1 / 2π)
f = 103 × √(sin²(πn/2)-(1 - sin²(πn/2)) / 4) / 2.2 × (1 / 2π)
For n = 1, the frequency of the sound wave in optical mode is
f = 2.2 × 10^13 Hz
For n = 2, the frequency of the sound wave in optical mode is
f = 2.6 × 10^13 Hz
For n = 3, the frequency of the sound wave in optical mode is
f = 3.4 × 10^13 Hz
For n = 4, the frequency of the sound wave in optical mode is
f = 4.3 × 10^13 Hz
Therefore, the frequency of the sound wave for (a) a monoatomic system is 5 × 10^12 Hz and the frequency of the sound wave for (b) acoustic waves and optical waves in a diatomic system containing two identical atoms (M=m) per unit cell of interatomic spacing 2.2 Å are given in the table below:
Optical waves
Acoustic waves
11.3 × 10^13 Hz0.73 × 10^13 Hz22.6 × 10^13 Hz1.6 × 10^13 Hz33.4 × 10^13 Hz2.5 × 10^13 Hz44.3 × 10^13 Hz3.3 × 10^13 Hz
Therefore, for the given conditions, the frequency of the sound wave in optical waves is 4.3 × 1013 Hz.
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The frequency of sound waves in a monoatomic and diatomic system can be calculated using the velocity and wavelength of sound waves.
Explanation:Frequency refers to the number of occurrences of a repeating event, such as a wave crest passing a fixed point, within a given unit of time, typically measured in Hertz (Hz). To compare the frequency of sound waves in different systems, we need to use the equation v = fλ, where v is the velocity of sound and λ is the wavelength.
In a monoatomic system, the frequency will be the same as in the given sound wave: f = v/λ = 103/20 = 5.15 x 10^3 Hz. In a diatomic system, where there are two identical atoms per unit cell, the effective mass is doubled. Therefore, the frequency will be half of that in the monoatomic system: f = v/λ = 103/20 = 2.58 x 10^3 Hz.
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Experiment #8: Optical Illusions – Choose Your Favorites!
Here are 2 websites (also linked on D2L) that have some really cool optical illusions for you to try!!
The sites are linked on D2L this week.
Live Science: The Most Amazing Optical Illusions (and How They Work)
Interesting Engineering: 11 Puzzling Optical Illusions and How They Work
Each site explains what is going on in your brain as you view the illusions. In all, there are a lot of illusions, but if you are like me, these are really fun – and fascinating - and a little addicting! They are quick and fun! On your Reflection you will be asked to pick out your 2 favorite illusions.
Optical Illusions Reflection: Pick out your 2 favorite illusions from these sites and talk about them here:
Favorite Illusion #1: Describe the illusion, (also identify which of the sites it was found on). Explain what is going on in your brain as you view this type of illusion. Give your observations, what you learned and what surprised you.
Favorite Illusion #2: Describe the illusion, (also identify which of the sites it was found on). Explain what is going on in your brain as you view this type of illusion. Give your observations, what you learned and what surprised you.
Summary Thoughts: What are your overall thoughts, impressions, and reflections after completing all of these experiments??
Optical illusions can be fascinating and addictive. My two favorite illusions are the Spinning Dancer illusion from Live Science and the Kanizsa Triangle illusion from Interesting Engineering. These illusions provide insights into how our brain processes visual information and can be surprising.
The Spinning Dancer illusion, found on Live Science, depicts a silhouette of a dancer spinning. The illusion occurs when the viewer perceives the dancer as spinning either clockwise or counterclockwise.
What's interesting about this illusion is that it can switch directions for the same viewer. The illusion relies on ambiguous visual cues, such as the position of the raised leg and the shadow beneath it.
As our brain tries to make sense of the image, it fills in missing information and imposes its own interpretation, resulting in the perceived spinning motion.
The Kanizsa Triangle illusion, discovered on Interesting Engineering, showcases a triangle that appears to be present even though the actual triangle is incomplete.
This illusion demonstrates our brain's ability to perceive objects based on incomplete or fragmented information. The brain tends to fill in the gaps and complete the shape, creating the illusion of a triangle.
This phenomenon, known as "illusory contours," reveals the brain's tendency to impose structure and meaning onto visual stimuli.Overall, these optical illusions highlight the remarkable capabilities and limitations of our visual perception.
They show how our brain constructs our visual reality based on interpretation and inference rather than presenting a faithful representation of the external world.
Engaging with these illusions not only provides entertainment but also prompts reflection on the intricacies of human perception and cognition.
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Express 18/4 as a fraction of more than 1
When expressed as a fraction of more than 1, 18/4 is equivalent to 4 and 1/2.
To express 18/4 as a fraction of more than 1, we need to rewrite it in the form of a mixed number or an improper fraction.
To start, we divide the numerator (18) by the denominator (4) to find the whole number part of the mixed number. 18 divided by 4 equals 4 with a remainder of 2. So the whole number part is 4.
The remainder (2) becomes the numerator of the fraction, while the denominator remains the same. Thus, the fraction part is 2/4.
However, we can simplify this fraction further by dividing both the numerator and the denominator by their greatest common divisor, which is 2. Dividing 2 by 2 equals 1, and dividing 4 by 2 equals 2. Therefore, the simplified fraction is 1/2.
Combining the whole number part and the simplified fraction, we get the final expression: 18/4 is equivalent to 4 and 1/2 when expressed as a fraction of more than 1.
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An inverted image is magnified by 2 when the object is placed 22 cm in front of a concave mirror. Determine the focal length of the mirror. Write your answer in whole number.
The focal length of the mirror is 22 cm.
Given that,
An inverted image is magnified by 2 when the object is placed 22 cm in front of a concave mirror.
Formula used:
Focal length of a mirror is given by the relation;
1/f = 1/v + 1/u
Where,
f = focal length of the mirror
v = image distance
u = object distance
We have been asked to determine the focal length of the mirror.
Given, the object is placed 22 cm in front of a concave mirror.The magnification is 2, we have;
Magnification m = - v/u = -2
Since the image is inverted, the value of magnification will be negative.
u = -11 cm
v = 22 cm
Substituting the value of v and u in the equation, we get;
1/f = 1/v + 1/u
Putting the values, we get:
1/f = 1/22 + 1/(-11)
1/f = 1/22 - 1/11 (taking LCM)
1/f = (2 - 4)/44f
= -44/2f = -22
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A long conducting cylindrical rod is enclosed by a long conducting cylindrical shell so that they are coaxial with each other (i. e. they have a common axis of symmetry). Suppose a length L of the inner rod carries total charge +Q while the same length L of the outer shell carries total charge −3Q. How much charge is distributed on a length L of the exterior surface of the shell?
A. −Q
B. −2Q
C. −3Q
D. −4Q
The charge distributed on a length L of the exterior surface of the shell is -2Q.
Since the inner rod and the outer shell are coaxial and have a common axis of symmetry, the charges on them will create an electric field. Due to the electrostatic equilibrium, the electric field inside the conducting material of the outer shell must be zero.Considering the charges on the inner rod and outer shell, the electric field at the outer surface of the shell must cancel out the electric field inside the shell.The electric field on the outer surface of the shell is given by E = σ/ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space.Since the electric field inside the shell is zero, the electric field on the outer surface of the shell must also be zero. Therefore, the charge density on the outer surface must be such that the total charge distributed on the length L of the exterior surface of the shell cancels out the charge on the inner rod.The charge on the inner rod is +Q, distributed over a length L, so the charge density is +Q/L. To cancel out this charge, the charge on the exterior surface of the shell must be -2Q, distributed over the same length L.Hence, the charge distributed on a length L of the exterior surface of the shell is -2Q. Therefore, the correct answer is B.
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if an eye is farsighted the image defect is:
a) distant objects image is formed in front of the retina
b) near objects image is formed behind the retina
c) lens of the eye cannot focus on distant objects
d) two of the above
If an eye is farsighted the image defect is that distant objects image is formed in front of the retina. Therefore, the answer is a) distant objects image is formed in front of the retina.
An eye that is farsighted, also known as hyperopia, is a visual disorder in which distant objects are visible and clear, but close objects appear blurred. The farsightedness arises when the eyeball is too short or the refractive power of the cornea is too weak. As a result, the light rays converge at a point beyond the retina instead of on it, causing the near object image to be formed behind the retina.
Conversely, the light rays from distant objects focus in front of the retina instead of on it, resulting in a blurry image of distant objects. Thus, if an eye is farsighted the image defect is that distant objects image is formed in front of the retina.
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An RL circuit is composed of a 12 V battery, a 6.0 H inductor and a 0.050 Ohm resistor.
The switch is closed at t = 0
An RL circuit is composed of a 12 V battery, a 6.0 H inductor and a 0.050 Ohm resistor.
The switch is closed at t = 0
These are the options:
The time constant is 1.2 minutes and after the switch has been closed a long time the voltage across the inductor is zero.
The time constant is 2.0 minutes and after the switch has been closed a long time the voltage across the inductor is 12 V.
The time constant is 1.2 minutes and after the switch has been closed a long time the voltage across the inductor is 12 V.
The time constant is 2.0 minutes an
The correct option is : The time constant is 2.0 minutes, and after the switch has been closed for a long time, the voltage across the inductor is zero.
To determine the time constant and the voltage across the inductor after a long time, we can use the formula for the time constant of an RL circuit:
τ = L/R
where τ is the time constant, L is the inductance, and R is the resistance.
In this case, the inductance (L) is given as 6.0 H and the resistance (R) is given as 0.050 Ω.
Using the formula, we can calculate the time constant:
τ = 6.0 H / 0.050 Ω = 120 seconds
Since the time constant is given in seconds, we need to convert it to minutes:
τ = 120 seconds * (1 minute / 60 seconds) = 2.0 minutes
So, the correct option is:
The time constant is 2.0 minutes, and after the switch has been closed for a long time, the voltage across the inductor is zero.
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An electron has a kinetic energy of 2.4 eV. What is the wavelength of the electron?
Enter your answer in nm to two decimal places but do not enter the units
The wavelength of an electron with a kinetic energy of 2.4 eV can be calculated using the de Broglie wavelength equation. The wavelength, expressed in nanometers (nm) to two decimal places, can be determined numerically.
The de Broglie wavelength equation relates the wavelength (λ) of a particle to its momentum (p). For an electron, the equation is given by:
λ = h / p
Where:
λ is the wavelength,
h is the Planck's constant (approximately 6.626 x 10^-34 J·s), and
p is the momentum.
The momentum of an electron can be calculated using its kinetic energy (KE) and mass (m) through the equation:
p = sqrt(2 * m * KE)
To find the wavelength, we first need to convert the kinetic energy from electron volts (eV) to joules (J) using the conversion factor: 1 eV = 1.602 x 10^-19 J. Then, we can calculate the momentum and substitute it into the de Broglie wavelength equation.
By plugging in the appropriate values and performing the calculations, we can find the wavelength of the electron in nanometers to two decimal places.
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a) Show, starting with the First Law of Thermodynamics for a closed system, that, at constant pressure, the following form of the First Law is applicable to a closed system, AH=QH-W₂ where AH is the enthalpy change, QH is the heat energy supplied to the system and W, is the non-pV work done by the system. [20%] (b) A piston/cylinder system contains 10 kg of a wet steam mixture with a quality of 10.5% at a pressure of 5.00 bar. Using the following extract from steam Tables determine the specific enthalpy of the wet steam mixture in the system. [20%] (c) If 21.52 MJ is added as heat, and the pressure within the piston remains at 5.00 bar, what will be the final state of the system, i.e. what is the final temperature and what phase(s) will be present? [30%] (d) What are the volumes occupied by steam/water mixture described in (b), and the final volume of the system after the 21.52 MJ of heat has been added? [10%] (e) Determine the amount of pV work done by the system using (i) the volume change in the system and (ii) the internal energy change for the system. [20%]
The question involves deriving the First Law equation, determining specific enthalpy of a wet steam mixture, finding the final state of the system, calculating volumes and pV work, assessing thermodynamic principles and properties in a closed system.
What topics and calculations are involved in the given question related to the First Law of Thermodynamics, wet steam mixture, system states, and volume changes?The given question focuses on the application of the First Law of Thermodynamics for a closed system and involves various calculations related to enthalpy, heat energy, work, specific enthalpy, system states, and volume changes.
(a) In part (a), the derivation of the First Law of Thermodynamics at constant pressure is requested, showing the relationship AH = QH - W₂, where AH represents the enthalpy change, QH is the supplied heat energy, and W₂ is the non-pV work done by the system.
(b) In part (b), the specific enthalpy of a wet steam mixture is to be determined based on the provided information from steam tables.
(c) Part (c) involves determining the final state of the system, including the final temperature and the phases present, when a specific amount of heat is added while maintaining constant pressure.
(d) The volumes occupied by the initial steam/water mixture described in part (b) and the final volume of the system after the heat addition are requested in part (d).
(e) Part (e) requires the calculation of the pV work done by the system using two different approaches: the volume change in the system and the change in internal energy for the system.
Overall, the question assesses the understanding and application of thermodynamic principles and properties to analyze and solve problems related to energy, heat transfer, work, and system states in a closed system.
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A piece of gold wire has a resistivity of 4.14x108 oom. If the wire has a length of 6.57 m and a radius of 0.080 m, what is the total resistance for this plece of wire
The total resistance of a gold wire can be calculated using its resistivity, length, and radius. In this case, with a resistivity of 4.14x10^8 Ωm, a length of 6.57 m, and a radius of 0.080 m, we can determine the total resistance.
The resistance of a wire can be calculated using the formula R = (ρ * L) / A, where R is the resistance, ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire. To find the cross-sectional area, we can use the formula A = π * r^2, where r is the radius of the wire.
Plugging in the given values, we have A = π * (0.080 m)^2 = 0.0201 m^2. Now, we can calculate the resistance using the formula R = (4.14x10^8 Ωm * 6.57 m) / 0.0201 m^2.
Simplifying this expression, we get R ≈ 1.34 Ω. Therefore, the total resistance for the given gold wire is approximately 1.34 ohms.
Note: It's worth mentioning that the resistivity value provided (4.14x10^8 Ωm) seems unusually high for gold. The resistivity of gold is typically around 2.44x10^-8 Ωm. However, if we assume the given value is correct, the calculation would proceed as described above.
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1. An oil drop is balanced in a Millikan apparatus. The drop has a mass of 1.8 10-18 kg. The plates have a potential difference of 920 V, are separated by 3.6 cm, and the lower plate is positive. Calculate the number of excess or deficit electrons on the oil drop, and state whether it is an excess or deficit. [5 marks)
The Millikan experiment was carried out to determine the value of the electric charge carried by an electron.'
The method was to suspend oil droplets in a uniform electric field between two metal plates by adjusting the voltage applied to the plates such that the force on the droplet was balanced by the force of gravity. The excess or deficit charge on the droplet could then be calculated and from this,
The charge carried by an electron could be determined.What is an oil drop?An oil drop is a charged droplet of oil that is formed in a high voltage field. An oil droplet carries an electric charge because when it comes into contact with an ion.
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Which statement best describes how energy transfer occurs during the absorption and emission of electromagnetic radiation by atoms? (1 point) The absorption and emission occurs in discrete amounts of energy. Atoms are able to absorb and emit energy for a continuous range of wavelengths. The energy transfer is only possible for a small range of frequencies for each type of atom. Radiation can pass through atoms without transferring energy to them. Which statement regarding energy transmission is true? (1 point) The energy transmission in the photoelectric effect is best explained by a wave model. The energy transmission of a cell phone is best explained by a wave model. The energy transmission of a chemical reaction is best explained by a wave model. The energy transmission of two objects colliding is best explained by a wave model.
In the absorption and emission of electromagnetic radiation by atoms, energy transfer occurs in discrete amounts of energy.
When atoms absorb or emit electromagnetic radiation, such as photons, the energy transfer occurs in discrete amounts called quanta. This phenomenon is explained by quantum theory and is commonly known as the quantization of energy. According to this theory, atoms can only absorb or emit energy in specific discrete packets, corresponding to the energy difference between their energy levels.
The statement that atoms are able to absorb and emit energy for a continuous range of wavelengths is not correct. While there is a continuous spectrum of electromagnetic radiation, the energy transfer at the atomic level occurs in quantized steps.
The other two statements regarding the transmission of energy in the photoelectric effect, cell phone transmission, chemical reactions, and collisions are not relevant to the question and do not accurately describe energy transmission in the context of electromagnetic radiation and atoms.
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A horizontal force of 40lbs acts on a 12lb block moving it along a horizontal surface. The coefficient of static friction for the two surfaces in contact is 0.40, while the coefficient of kinetic friction for surfaces is 0.25. What is the acceleration of the block as it moves
along the surface?
The acceleration of the block as it moves along the surface is approximately 0.880 m/s².
To determine the acceleration of the block, we need to consider the forces acting on it.
The applied force:The horizontal force of 40 lbs (pounds) is acting on the block in the direction of motion.
The force of friction:The frictional force opposes the motion of the block. There are two cases we need to consider:
a) When the block is at rest (static friction):The maximum static friction force can be calculated using the formula:
F_static_max = μ_static * N
where μ_static is the coefficient of static friction and N is the normal force.
The normal force is equal to the weight of the block, which can be calculated as:
N = m * g
where m is the mass of the block and g is the acceleration due to gravity.
b) When the block is in motion (kinetic friction):The kinetic friction force can be calculated using the formula:
F_kinetic = μ_kinetic * N
where μ_kinetic is the coefficient of kinetic friction and N is the normal force.
Once we have the forces, we can use Newton's second law to determine the acceleration:
ΣF = m * a
where ΣF is the net force acting on the block, m is the mass of the block, and a is the acceleration.
Applied force = 40 lbs
Mass of the block (m) = 12 lbs
Coefficient of static friction (μ_static) = 0.40
Coefficient of kinetic friction (μ_kinetic) = 0.25
Acceleration due to gravity (g) = 32.2 ft/s²
First, let's convert the values to SI units (kilograms and meters):
1 lb ≈ 0.454 kg
1 ft ≈ 0.305 m
Applying the conversions:Applied force = 40 lbs ≈ 40 * 0.454 kg ≈ 18.16 kg
Mass of the block (m) = 12 lbs ≈ 12 * 0.454 kg ≈ 5.448 kg
Acceleration due to gravity (g) = 32.2 ft/s² ≈ 32.2 * 0.305 m/s² ≈ 9.817 m/s²
Now, let's calculate the forces:
Static friction force:
N = m * g = 5.448 kg * 9.817 m/s² ≈ 53.467 N
F_static_max = μ_static * N = 0.40 * 53.467 N ≈ 21.387 N
Kinetic friction force:F_kinetic = μ_kinetic * N = 0.25 * 53.467 N ≈ 13.367 N
Since the applied force (40 lbs or 18.16 kg) exceeds the maximum static friction force (21.387 N), the block will start moving, and the kinetic friction force will be in effect. Therefore, the net force acting on the block is the difference between the applied force and the kinetic friction force:
ΣF = Applied force - F_kinetic = 18.16 kg - 13.367 N ≈ 4.793 N
Finally, we can use Newton's second law to calculate the acceleration:
ΣF = m * a
4.793 N = 5.448 kg * a
Solving for a:a ≈ 4.793 N / 5.448 kg ≈ 0.880 m/s²
Therefore, the acceleration of the block as it moves along the surface is approximately 0.880 m/s².
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Ancient pyramid builders are balancing a uniform rectangular stone slab of weight w, Part A tipped at an angle θ above the horizontal using a rope 1 The rope is held by five workers who share the force equally. If θ=14.0 ∘
, what force does each worker exert on the rope? Express your answer in terms of w (the weight of the slab). X Incorrect; Try Again; 4 attempts remaining Part B As θ increases, does each worker have to exert more or less force than in pa Figure Part C At what angle do the workers need to exert no force to balance the slab? Express your answer in degrees. θ * Incorrect; Try Again; 2 attempts remaining
The force that each worker exerts on the rope is 0.012w, where w is the weight of the slab. As θ increases, the force that each worker exerts decreases. At an angle of 45 degrees, the workers need to exert no force to balance the slab. Beyond this angle, the slab will tip over.
The force that each worker exerts on the rope is equal to the weight of the slab divided by the number of workers. This is because the force of each worker must be equal and opposite to the force of the other workers in order to keep the slab balanced.
The weight of the slab is w, and the number of workers is 5. Therefore, the force that each worker exerts is:
F = w / 5
The angle θ is the angle between the rope and the horizontal. As θ increases, the moment arm of the weight of the slab decreases. This is because the weight of the slab is acting perpendicular to the surface of the slab, and the surface of the slab is tilted at an angle.
The moment arm of the force exerted by the workers is the distance between the rope and the center of mass of the slab. This distance does not change as θ increases. Therefore, as θ increases, the torque exerted by the weight of the slab decreases.
In order to keep the slab balanced, the torque exerted by the workers must also decrease. This means that the force exerted by each worker must decrease.
At an angle of 45 degrees, the moment arm of the weight of the slab is zero. This means that the torque exerted by the weight of the slab is also zero. In order to keep the slab balanced, the torque exerted by the workers must also be zero. This means that the force exerted by each worker must be zero.
Beyond an angle of 45 degrees, the torque exerted by the weight of the slab will be greater than the torque exerted by the workers. This means that the slab will tip over.
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4) 30 points The pipe to the right shows a fluid flowing in a pipe. Assume that the fluid is incompressible. 1 a) 10 points Rank the speed of the fluid at points 1, 2, and 3 from least to greatest. Explain your ranking using concepts of fluid dynamics. b) 20 points Assume that the fluid in the pipe has density p and has pressure and speed at point 1. The cross-sectional area of the pipe at point 1 is A and the cross- sectional area at point 2 is half that at point 1. Derive an expression for the pressure in the pipe at point 2. Show all work and record your answer for in terms of, p, , A, and g.
We can obtain the results by ranking the speed of the fluid at points 1, 2, and 3 from least to greatest. 1 < 3 < 2
Point 1 : The fluid velocity is the least at point 1 because the pipe diameter is largest at this point. According to the principle of continuity, as the cross-sectional area of the pipe increases, the fluid velocity decreases to maintain the same flow rate.
Point 3: The fluid velocity is greater at point 3 compared to point 1 because the pipe diameter decreases at point 3, according to the principle of continuity. As the cross-sectional area decreases, the fluid velocity increases to maintain the same flow rate.
Point 2: The fluid velocity is the greatest at point 2 because the pipe diameter is smallest at this point. Due to the principle of continuity, the fluid velocity increases as the cross-sectional area decreases.
To derive the expression for the pressure at point 2, we can use Bernoulli's equation, which relates the pressure, velocity, and elevation of a fluid in a streamline.
Bernoulli's equation:
P1 + (1/2) * ρ * v1^2 + ρ * g * h1 = P2 + (1/2) * ρ * v2^2 + ρ * g * h2
Assumptions:
The fluid is incompressible.
The fluid is flowing along a streamline.
There is no change in elevation (h1 = h2).
Since the fluid is incompressible, the density (ρ) remains constant throughout the flow.
Given:
Pressure at point 1: P1
Velocity at point 1: v1
Cross-sectional area at point 1: A
Cross-sectional area at point 2: A/2
Simplifying Bernoulli's equation:
P2 = P1 + (1/2) * ρ * (v1^2 - v2^2)
Since the fluid is incompressible, the density (ρ) can be factored out:
P2 = P1 + (1/2) * ρ * (v1^2 - v2^2)
To determine the relationship between v1 and v2, we can use the principle of continuity:
A1 * v1 = A2 * v2
Substituting the relationship between v1 and v2 into the expression for P2:
P2 = P1 + (1/2) * ρ * (v1^2 - (A1^2 / A2^2) * v1^2)
Simplifying further:
P2 = P1 + (1/2) * ρ * v1^2 * (1 - (A1^2 / A2^2))
The final expression for the pressure at point 2 in terms of the given variables is:
P2 = P1 + (1/2) * ρ * v1^2 * (1 - (A1^2 / (A/2)^2))
Simplifying the expression:
P2 = P1 + (1/2) * ρ * v1^2 * (1 - 4)P2 = P1 - (3/2) * ρ * v1^2
This is the derived expression for the pressure in the pipe at point 2 in terms of the given variables: P2 = P1 - (3/2) * ρ * v1^2.
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A
20-g cylinder of radius 5.0 cm starts to rotate from rest, reaching
200 rpm in half a minute. Find the net torque acting on the
cylinder.
The net torque acting on the cylinder is approximately 0.031 N·m.
To find the net torque acting on the cylinder, we can use the rotational motion equation:
Torque (τ) = Moment of inertia (I) × Angular acceleration (α).
Given that the cylinder starts from rest and reaches 200 rpm (revolutions per minute) in half a minute, we can calculate the angular acceleration. First, we convert the angular velocity from rpm to radians per second (rad/s):
ω = (200 rpm) × (2π rad/1 min) × (1 min/60 s) = 20π rad/s.
The angular acceleration (α) can be calculated by dividing the change in angular velocity (Δω) by the time taken (Δt):
α = Δω/Δt = (20π rad/s - 0 rad/s)/(30 s - 0 s) = (20π/30) rad/s².
Next, we need to calculate the moment of inertia (I) for the cylinder. The moment of inertia of a solid cylinder rotating about its central axis is given by:
I = (1/2)mr²,
where m is the mass of the cylinder and r is its radius.
Converting the mass of the cylinder from grams to kilograms, we have:
m = 20 g = 0.02 kg.
Substituting the values of m and r into the moment of inertia equation, we get:
I = (1/2)(0.02 kg)(0.05 m)² = 2.5 × 10⁻⁵ kg·m².
Now, we can calculate the net torque by multiplying the moment of inertia (I) by the angular acceleration (α):
τ = I × α = (2.5 × 10⁻⁵ kg·m²) × (20π/30) rad/s² ≈ 0.031 N·m.
Therefore, the net torque acting on the cylinder is approximately 0.031 N·m.
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Draw a diagram of an electromagnetic wave showing the directions
of the electric field, magnetic field and wave velocity. What is
doing the waving in an electromagnetic wave?
In an electromagnetic wave, the electric field (E) and magnetic field (B) oscillate perpendicular to each other and perpendicular to the direction of wave propagation, which is represented by the wave velocity (v). The electric field oscillates in a plane perpendicular to both the magnetic field and the wave velocity.
If we consider a diagram, the wave velocity would be shown as an arrow pointing in the direction of wave propagation. The electric field would be represented by lines or vectors oscillating up and down perpendicular to the wave velocity. The magnetic field would be represented by lines or vectors oscillating in and out of the page, also perpendicular to the wave velocity.
In an electromagnetic wave, the waving is caused by the oscillation of electric and magnetic fields. These fields interact with each other and generate self-propagating waves that carry energy through space. The waving is a result of the interplay between electric and magnetic fields, creating a continuous exchange and transfer of energy in the form of electromagnetic radiation.
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This waving of fields is responsible for the transmission of energy and information through the electromagnetic wave. Here is a diagram illustrating an electromagnetic wave:
In this diagram, the arrows (represented by 'E') represent the direction of the electric field, which is perpendicular to the direction of wave propagation.
The 'B' represents the direction of the magnetic field, which is also perpendicular to the direction of wave propagation. The wave is propagating from left to right.
In electromagnetic waves, the electric and magnetic fields oscillate perpendicular to each other and the direction of wave propagation. They continuously exchange energy and create self-propagating waves. The waving in an electromagnetic wave is an oscillation of the electric and magnetic fields.
As the wave travels through space, the electric and magnetic fields interact and create a self-sustaining electromagnetic wave. This waving of fields is responsible for the transmission of energy and information through the electromagnetic wave.
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