Inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. L is used to represent the inductance, and Henry is the SI unit of inductance. The total inductance of the series connection with opposing fields and a mutual inductance of 0.2H is 0.71H.
The resonant frequency of the RC circuit with a 470-ohm resistor and a 120 uF capacitor is 2.82 Hz.
10. When inductances are connected in series with opposing fields, the total inductance is given by the difference between the individual inductances:
total inductance = (L1 + L2) - 2M
Given the values of L1 = 0.8H, L2 = 0.11H, and M = 0.2H, we can substitute these values into the formula to find the total inductance, which is 0.71H.
11. The resonant frequency of an RC circuit is determined by the values of the resistor and capacitor. The resonant frequency is given by the formula:
resonant frequency = 1 / (2πRC)
Plugging in the values of R = 470 ohms and C = 120 uF (or 120 × 10^(-6) F), we can calculate the resonant frequency, which is 2.82 Hz.
It's important to note that in the calculations above, I've assumed ideal conditions and neglected any parasitic elements or losses in the circuit. In practical scenarios, these factors may affect the actual values and performance of the components.
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The oscillation of a spring block with mass of 5 kg is described by the equation y = 0.10m cos (2nt). What is the spring constant k? 15.7 N/m O 197 N/m O 31.4 N/m O 5.0 N/m
The oscillation of a spring block with mass of 5 kg is described by the equation y = 0.10m cos (2nt). So the spring constant (k) for the given oscillation equation is 5.0 N/m.
In the equation y = A * cos(2πnt), where y is the displacement, A is the amplitude, n is the frequency, and t is time, we can see that the angular frequency (ω) is given by 2πn.
Comparing this with the equation for simple harmonic motion, y = Acos(ωt), we can see that the angular frequency ω is related to the spring constant k and the mass m by the equation ω = √(k/m).
In our given equation, we have ω = 2πn. Since we know the mass of the block is 5 kg, we can solve for k.
k = mω² = (5 kg) * (2πn)² = 5 * 4π²n² = 5 * (39.48n²) = 197.4n².
Therefore, the spring constant k is 197.4n² N/m.
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A hiker walks 300ft 15 degrees north of west and 0.7 km northeast. Calculate the magnitude and direction of the hiker's total displacement.
To calculate the magnitude and direction of the hiker's total displacement, we need to find the vector sum of the two displacement vectors: 300 ft at 15 degrees north of west and 0.7 km northeast.
First, we need to convert the given distances to a common unit. Let's convert 300 ft to kilometers. Since 1 km is equal to 3280.84 ft, the conversion is: 300 ft × (1 km / 3280.84 ft) = 0.09144 km.
Next, we can represent the first displacement vector as V1 = 0.09144 km at 15 degrees north of west. To simplify calculations, we can break it down into its horizontal (west) and vertical (north) components using trigonometry. The horizontal component (V1x) is given by V1x = 0.09144 km × cos(15 degrees), and the vertical component (V1y) is given by V1y = 0.09144 km × sin(15 degrees).
Now, let's consider the second displacement vector, which is 0.7 km northeast. To determine its horizontal (V2x) and vertical (V2y) components, we can use the fact that northeast is a 45-degree angle between north and east. Therefore, V2x = 0.7 km × cos(45 degrees) and V2y = 0.7 km × sin(45 degrees).
To find the total displacement, we can sum the horizontal and vertical components: Vx = V1x + V2x and Vy = V1y + V2y.
Finally, the magnitude of the total displacement (V) can be calculated using the Pythagorean theorem: V = sqrt(Vx^2 + Vy^2). The direction of the total displacement can be determined using the inverse tangent function: direction = atan(Vy / Vx).
By substituting the values and performing the calculations, we can determine the magnitude and direction of the hiker's total displacement.
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Any two point charges exert equally strong electric forces on each other. Coulomb's constant is 8.99 x 10° N-m²/C², and given that an electron has a charge of -1.60 x 10 19 C What is the electric force (magnitude and direction) between two electrons (-e) separated by a distance of 15.5 cm?
The impulse delivered to the tennis ball is 0.54 N-s, Impulse is defined as the product of force and time. In this case, the force is 180 N and the time interval is 0.005-0.002 = 0.003 seconds.
Therefore, the impulse delivered to the tennis ball is: Impulse = Force * Time = 180 N * 0.003 seconds = 0.54 N-s
The impulse delivered to the tennis ball is a measure of the change in momentum of the ball. In this case, the impulse causes the ball to accelerate from rest to a velocity of 180 N / 0.54 N-s = 333 m/s.
The impulse delivered to the tennis ball can also be calculated by using the area under the force-time curve. In this case, the area under the curve is a triangle with a base of 0.003 seconds and a height of 180 N. Therefore, the impulse delivered to the tennis ball is:
Impulse = Area under the force-time curve = (1/2) * Base * Height = (1/2) * 0.003 seconds * 180 N = 0.54 N-s
Therefore, the impulse delivered to the tennis ball is 0.54 N-s.
We first calculate the area under the force-time magnitude curve. This is done by multiplying the base of the triangle by the height of the triangle.
We then divide the area by 2, since the area of a triangle is equal to (1/2) * base * height. This gives us the impulse delivered to the tennis ball.
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Maximize points by completing the tasks correctly on first attempt. Incorrect responses incur a 2 point penalty per attempt. Calculate the distance and luminosity of Dargo in the following units: Distance meters Luminosity watts L s
Calculate the luminosities, in L s
, of the other stars.
No, without additional information such as apparent magnitude or spectral type, it is not possible to accurately calculate the distance and luminosity of Dargo or the luminosities of other stars.
Can the distance and luminosity of Dargo be calculated without additional information?In order to calculate the distance and luminosity of Dargo, we need additional information such as the star's apparent magnitude, spectral type, or any other relevant data. Without this information, it is not possible to provide accurate calculations for the distance and luminosity of Dargo or the luminosities of other stars.
Distance in astronomy is typically measured using units such as parsecs (pc) or light-years (ly). Luminosity, on the other hand, is a measure of the total amount of energy emitted by a star per unit time and is usually expressed in units of watts (W) or solar luminosities (L☉).
To calculate the distance to a star, methods such as parallax measurements or spectroscopic parallax can be used. These methods rely on observations and measurements of the star's apparent position or characteristics to determine its distance from Earth.
Luminosity can be calculated using various methods, including the star's temperature, radius, and the Stefan-Boltzmann law, which relates the luminosity of a star to its temperature and radius.
Without specific data or parameters for Dargo or other stars, it is not possible to provide accurate calculations for their distances or luminosities.
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In New England, the horizontal component of the earth's magnetic field has a magnitude of
1.6 x 10^-5 T. An electron is shot vertically straight up from the ground with a speed of
2.1 x 10^6 m/S What is the magnitude of the acceleration caused by the magnetic force? Ignore
the gravitational force acting on the electron.
The magnitude of the acceleration caused by the magnetic force on the electron is 6.72 x [tex]10^5[/tex] m/s^2.
The magnetic force on a moving charged particle is given by the formula:
F = qvB
where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field.
In this case, the electron has a charge of -1.6 x [tex]10^-19[/tex] C (negative because it's an electron), a velocity of 2.1 x [tex]10^6[/tex] m/s, and the magnitude of the horizontal component of the Earth's magnetic field is 1.6 x [tex]10^-5[/tex] T.
Substituting the values into the formula, we have:
F = (-1.6 x [tex]10^-19[/tex]) * (2.1 x [tex]10^6[/tex]) * (1.6 x [tex]10^-5[/tex])
≈ -5.376 x [tex]10^-18[/tex] N
Since the force is acting in the opposite direction to the motion of the electron, we take its magnitude:
|F| = 5.376 x [tex]10^-18[/tex] N
The acceleration caused by the magnetic force can be calculated using Newton's second law:
F = ma
Rearranging the formula, we have:
a = F / m
The mass of the electron is approximately 9.11 x [tex]10^-31[/tex] kg. Substituting the values, we get:
a = (5.376 x [tex]10^-18[/tex]) / (9.11 x[tex]10^-31[/tex])
≈ 6.72 x[tex]10^5[/tex] m/s^2
Therefore, the magnitude of the acceleration caused by the magnetic force on the electron is approximately 6.72 x [tex]10^5[/tex] m/s^2.
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Consider a wire loop of radius r and at a distance of L from a long straight current carrying conductor in the plane of loop. Under what conditions that we can write the flux through the loop as Φ=( 2πr
μ 0
I
)πa 2
Calculate the mutual inductance if the loop radius is 1 cm and the distance from the straight conductor is 0.5 m.
To calculate the mutual inductance between a wire loop and a long straight current-carrying conductor, we can use the formula Φ = (2πrμ₀I)πa²,
where Φ is the flux through the loop, r is the radius of the loop, μ₀ is the permeability of free space, I is the current in the straight conductor, and a is the distance between the loop and the conductor.
In this case, the loop radius is given as 1 cm (or 0.01 m) and the distance from the straight conductor is 0.5 m. By substituting these values into the formula, we can find the mutual inductance.
Φ = (2π(0.01 m)(4π×10^(-7) T·m/A)(I))π(0.5 m)²
Φ = 4π²(0.01)(4π×10^(-7) I)×(0.5)²
Φ = 0.004π²I
Therefore, the mutual inductance is given by 0.004π² times the current in the straight conductor.
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A 50 kg sprinter, starting from rest, runs 43 m in 7.0 s at constant acceleration. a) What is the magnitude of the horizontal force acting on the sprinter? Express your answer with the appropriate units. b) What is the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s? Express your answers in kilowatts separated by commas.
2) A pendulum is made by tying a 490 g ball to a 42.0 cm long string. The pendulum is pulled 25.0 degrees to one side, then released. a) What is the ball's speed at the lowest point of its trajectory? Express your answer with the appropriate units. b) To what angle does the pendulum swing on the other side? Express your answer with the appropriate units.
a) the magnitude of the horizontal force acting on the sprinter is approximately 54.5 N.
b) power output can be calculated using the equation: \[P = \frac{W}{t}\]
a) To find the magnitude of the horizontal force acting on the sprinter, we can use the equation of motion for constant acceleration:
\[d = \frac{1}{2}at^2\]
where \(d\) is the distance traveled, \(a\) is the acceleration, and \(t\) is the time.
Given: \(d = 43 \, \text{m}\), \(t = 7.0 \, \text{s}\)
We can rearrange the equation to solve for acceleration:
\[a = \frac{2d}{t^2}\]
Substituting the given values:
\[a = \frac{2 \times 43 \, \text{m}}{(7.0 \, \text{s})^2}\]
\[a \approx 1.09 \, \text{m/s}^2\]
The horizontal force acting on the sprinter can be found using Newton's second law:
\[F = ma\]
Given: \(m = 50 \, \text{kg}\), \(a = 1.09 \, \text{m/s}^2\)
\[F = (50 \, \text{kg})(1.09 \, \text{m/s}^2)\]
\[F \approx 54.5 \, \text{N}\]
Therefore, the magnitude of the horizontal force acting on the sprinter is approximately 54.5 N.
b) Power is defined as the rate at which work is done or energy is transferred. The power output can be calculated using the equation:
\[P = \frac{W}{t}\]
where \(P\) is the power, \(W\) is the work done, and \(t\) is the time.
In this case, the work done can be calculated using the equation:
\[W = \frac{1}{2}mv^2\]
Given: \(m = 50 \, \text{kg}\), \(v\) can be calculated using the equation for uniformly accelerated motion:
\[v = u + at\]
where \(u\) is the initial velocity (which is 0 since the sprinter starts from rest), \(a\) is the acceleration (which is the same as calculated in part a), and \(t\) is the time.
For \(t = 2.0 \, \text{s}\):
\[v = (0) + (1.09 \, \text{m/s}^2)(2.0 \, \text{s})\]
\[v = 2.18 \, \text{m/s}\]
For \(t = 4.0 \, \text{s}\):
\[v = (0) + (1.09 \, \text{m/s}^2)(4.0 \, \text{s})\]
\[v = 4.36 \, \text{m/s}\]
For \(t = 6.0 \, \text{s}\):
\[v = (0) + (1.09 \, \text{m/s}^2)(6.0 \, \text{s})\]
\[v = 6.54 \, \text{m/s}\]
Now we can calculate the power output:
For \(t = 2.0 \, \text{s}\):
\[P = \frac{1}{2}(50 \, \text{kg})(2.18 \, \text{m/s})^2\]
\[P \approx 119 \, \text{W}\]
For \(t = 4.0 \, \text{s}\):
\[P = \frac{1}{2}(50 \, \text
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(a) What length of pipe open at both ends has a fundamental frequency of 3.73 x 102 Hz? Find the first overtone. (pipe = 0.46 m Jovertone = 745.65 Hz (b) If the one end of this pipe is now closed, what is the new fundamental frequency? Find the first overtone. Stundamental = 186.41 ✓ Hz Sovertone = 559.24 Hz (c) If the pipe is open at one end only, how many harmonics are possible in the normal hearing range from 20 to 20000 Hz? n = 107.3 X What is the index n of the largest audible harmonic? Which harmonics are missing? How many sarmonics are audible?
The fundamental frequency of a pipe open at both ends is given by the equation f = (v/2L), where f is the frequency, v is the speed of sound, and L is the length of the pipe. Rearranging the equation, we can solve for L:
L = v / (2f)
Substituting the given values, with v being the speed of sound in air (approximately 343 m/s), and f being 3.73 x 10^2 Hz, we find:
L = 343 m/s / (2 * 3.73 x 10^2 Hz) ≈ 0.46 m
So, the length of the pipe open at both ends is approximately 0.46 meters.
For the first overtone, we can use the equation f1 = 2f, where f1 is the frequency of the first overtone. Substituting the given fundamental frequency (3.73 x 10^2 Hz), we find:
f1 = 2 * (3.73 x 10^2 Hz) = 7.46 x 10^2 Hz
Therefore, the frequency of the first overtone is approximately 7.46 x 10^2 Hz (or 745.65 Hz as given).
When one end of the pipe is closed, the fundamental frequency changes. In this case, the new fundamental frequency can be found using the equation f = v / (4L), where L is the length of the pipe. Substituting the given length (0.46 m) and the speed of sound in air, we find:
f = 343 m/s / (4 * 0.46 m) ≈ 186.41 Hz
So, the new fundamental frequency of the closed pipe is approximately 186.41 Hz.
To find the first overtone, we can use the equation f1 = 3f, where f1 is the frequency of the first overtone. Substituting the new fundamental frequency (186.41 Hz), we find:
f1 = 3 * 186.41 Hz ≈ 559.24 Hz
Therefore, the frequency of the first overtone in the closed pipe is approximately 559.24 Hz.
For a pipe open at one end only, the possible harmonics in the normal hearing range from 20 Hz to 20,000 Hz can be determined using the equation f = (2n - 1) * v / (4L), where n is the harmonic number. We can rearrange this equation to solve for n:
n = (4Lf) / v + 1/2
Substituting the given values of L (0.46 m) and v (343 m/s), we find:
n = (4 * 0.46 m * 20,000 Hz) / 343 m/s + 1/2 ≈ 107.3
Therefore, there are approximately 107 harmonics possible in the normal hearing range from 20 to 20,000 Hz for a pipe open at one end only.
The largest audible harmonic corresponds to the index n = 107, as calculated above. The missing harmonics are the ones that fall outside the audible range of 20 to 20,000 Hz. Since the fundamental frequency of the pipe is given by f = v / (4L), we can find the highest possible frequency (20,000 Hz) using this equation and solving for L:
L = v / (4f) = 343 m/s / (4 * 20,000 Hz) ≈ 0.0043 m
Any harmonics with a wavelength shorter than this length (or a frequency higher than 20,000 Hz)
would be outside the audible range. Therefore, the highest audible harmonic would be the one corresponding to the index n = 107.
the length of the pipe open at both ends with a fundamental frequency of 3.73 x 10^2 Hz is approximately 0.46 meters. The first overtone for this pipe has a frequency of 7.46 x 10^2 Hz (or 745.65 Hz). When one end of the pipe is closed, the new fundamental frequency becomes approximately 186.41 Hz, and the first overtone has a frequency of approximately 559.24 Hz. For a pipe open at one end only, there are approximately 107 harmonics possible in the normal hearing range from 20 to 20,000 Hz. The largest audible harmonic corresponds to the index n = 107, and the missing harmonics are the ones that fall outside the audible range.
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Part A To what temperature will 7600 J of heat raise 3.0 kg of water that is initially at 16.0°C? The specific heat of water is 4186 J/kg · Cº. Express your answer using three significant figures.
The temperature will raise by 0.605 °C on supplying 7600 J of heat to 3.0 kg of water that is initially at 16.0°C.
From the question above, Mass of water = 3.0 kg
Initial temperature of water = 16.0 °C
Specific heat of water = 4186 J/kg · °C
Heat supplied = 7600
Formula to calculate the change in temperature of the substance due to heat
Q = mcΔT
Here, Q is the heat supplied, m is the mass of the substance, c is the specific heat capacity of the substance and ΔT is the change in temperature of the substance on receiving the given amount of heat.
Supplying the given values,
7600 = 3.0 × 4186 × ΔT
ΔT = 7600 / (3.0 × 4186) = 0.6053...°C
Rounding off the above answer to three significant figures, we get
ΔT = 0.605 °C (approx)
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What is the instrument used to monitor swelling of a volcanoes flanks?
The instrument used to monitor swelling of a volcanoes flanks is a tiltmeter.
What is a volcano?In Science, a volcano can be defined as a cone-shaped landform that is typically formed through repeated eruptions over a period of time.
Additionally, a volcano simply refers to an opening that is typically formed within the Earth's crust through which ash, lava, and gases flow during an eruption.
A tiltmeter can be defined as a sensitive device that is designed and developed for the measurement of changes in the slope (rise and run) or tilt of the ground surface.
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3 5 points In the balanced three phase AC circuit in Figure 4, the Y-connected phasor voltage source has an a-b-c sequence with Van=100/15° V and the load impedance in each A- connected phase is ZA-100/45° N. a IAB ZA The phasor current IAB is 0.865-j1.498 A 1.73+j0 A 0.865+j1.498 A 0-j1.73 A Next Previous C V V... "an n Von b Ib Figure 4 Zs B IBC ZA ICA C --D
The phasor current IAB in the given balanced three-phase AC circuit is 0.865+j1.498 A.
What is the phasor current IAB in a balanced three-phase AC circuit with a Y-connected voltage source and A-connected load impedances?In the given balanced three-phase AC circuit, the Y-connected phasor
voltage source has a sequence of a-b-c, with the phase voltage Van being
100/15° V. Each phase of the A-connected load impedance ZA is
represented as ZA-100/45° Ω.
The question asks for the phasor current IAB. Based on the provided
options, the correct answer is 0.865+j1.498 A.
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Two blocks, m1=2 kg and m2=6 kg, are connected by a string passing over a pulley of radius r=0.25 m and moment of inertia I, as in the figure. If the magnitude of the linear acceleration of the system a=2 m/s2, then the moment of inertia (in kg.m²) of the pulley is: at m m2 O a. 0.333 O b. 0.125 ос. 0.5 O d. 0.75 O e. 2
The moment of inertia of the pulley is 0.5 kg·m², In this system, we have two blocks connected by a string passing over a pulley.
The blocks have masses of m1 = 2 kg and m2 = 6 kg, and the linear acceleration of the system is given as a = 2 m/s².
The moment of inertia of the pulley, we can use the principle of rotational motion. The net torque acting on the pulley is equal to the product of the moment of inertia and the angular acceleration.
Considering the forces acting on the system, we have the tension T in the string pulling block m1 upward and the weight mg of block m2 pulling it downward. The net torque is given by the difference in torque due to these forces.
The torque due to the tension T can be calculated as T * r, where r is the radius of the pulley. The torque due to the weight mg can be calculated as (m2 * g) * r.
Since the system is in equilibrium, the net torque is zero. Therefore, we can equate the torque due to tension and the torque due to weight: T * r = (m2 * g) * r.
From this equation, we can solve for the tension T, which is equal to (m2 * g).
Finally, we can use the equation for linear acceleration a = (m2 * g - T) / (m1 + m2) and substitute the values to find the acceleration.
That a = 2 m/s², m1 = 2 kg, m2 = 6 kg, and g = 9.8 m/s², we can solve for the tension T.
Using the obtained tension, we can calculate the moment of inertia I of the pulley using the equation I = (m1 * a * r - T * r) / (a * r²).
By substituting the known values, we find that I = 0.5 kg·m².
Therefore, the moment of inertia of the pulley is 0.5 kg·m².
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A 0.0450 kg loe cube at -30.0°C is placed in 0.477 kg of 35.0°C water in a very well Insulated container. What is the final temperature? The latent heat of fusion of water is 79.8 kcal/kg, the specific heat of ice is 0.50 kcal/(kg "C), and the specific heat of water is 1.00 kcal/(kg C). 24.9 x How many different changes are involved for the ice to go from -30.0°C to melted water at some final temperature? How is the quantity of heat gained by the ice related to the quantity of heat lost by the water C ,During a marathon race David uses energy at a rate of 272 W. What volume of body fluid does he lose in the 5.5 hours of the race if 21.0% of the energy goes to the muscle tissue and the rest is used in removing the perspiration from the body. The latent heat of vaporization is 2.41 x 10 /kg at 37.0°C and density of water is 1000 kg/m³ 1.8 x How is the quantity of heat during phase change related to the latent heat of vaporization?
1.There are four different changes involved for the ice to go from -30.0°C to melted water at some final temperature.
2.The quantity of heat gained by the ice is equal to the quantity of heat lost by the water.
3.To determine the final temperature of the system, we need to consider the heat transfer between the ice cube and the water. The heat gained by the ice cube is equal to the heat lost by the water.
The process involves four different changes for the ice cube: (1) heating the ice from -30.0°C to 0°C, (2) melting the ice at 0°C, (3) heating the water from 0°C to the final temperature, and (4) bringing the water to the final temperature.
1.During the first change, the ice cube gains heat as it is heated from -30.0°C to 0°C. This can be calculated using the specific heat of ice and the mass of the ice cube.
2.During the second change, the ice cube absorbs the latent heat of fusion as it melts into water at 0°C. The quantity of heat absorbed can be calculated using the latent heat of fusion and the mass of the ice cube.
3.During the third change, the water is heated from 0°C to the final temperature. The quantity of heat gained by the water can be calculated using the specific heat of water and the mass of the water.
Finally, during the fourth change, the water reaches the final temperature. The final temperature can be determined by equating the heat gained by the water to the total heat lost by the ice cube.
Regarding the second question, during a phase change (such as the melting of ice or the vaporization of water), the quantity of heat transferred is related to the latent heat of that specific phase change.
The latent heat of vaporization, for example, represents the amount of heat required to convert a substance from liquid to vapor at a constant temperature. It is the quantity of heat absorbed or released during the phase change process, without a change in temperature.
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A single phase bridge rectifier is supplied by V, = 220 V (rms) AC voltage. Load is a battery of which voltage is E = 200 V (DC). In order to limit the battery charging current, there is a seriesly connected resistance R = 10 Ohm to the battery. a. Draw input and output currents waveforms of the rectifiesr. b. Calculate average and rms battery charging current. c. Calculate Power Factor and Crest Factor of the rectifier.
The series resistance limits the battery charging current and provides current regulation in the single-phase bridge rectifier circuit.
What is the role of the series resistance in the single-phase bridge rectifier circuit with a battery load?a. The input current waveform of a single-phase bridge rectifier is pulsating and non-sinusoidal, while the output current waveform has a smoother DC component due to filtering.
b. Average and RMS battery charging current are approximately equal and can be calculated as E/R, where E is the battery voltage and R is the series resistance.
c. Power Factor (PF) is the ratio of average power to apparent power, and Crest Factor (CF) is the ratio of peak current to RMS current, both of which depend on the specific waveform and load characteristics.
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Another galaxy is receding from Earth with an unknown speed. The orange light emitted at a frequency of 5.000×1014Hz5.000×1014Hz by the galaxy, when reaching the Earth, is observed at a frequency of 4.9283×1014 HzHz . What is the receding speed of this galaxy in km/s?
The receding speed of the galaxy can be calculated using the redshift formula. In this case, the observed frequency of light and the emitted frequency of light are given, allowing us to determine the velocity. The receding speed of the galaxy is approximately 46,185 km/s.
The observed frequency of light, f_obs, and the emitted frequency of light, f_em, are related to the receding speed of the galaxy, v, through the redshift formula:
[tex]f_obs[/tex] = (1 + (v/c)) * f_em
where c is the speed of light. Rearranging the formula to solve for v, we have:
v = ([tex]f_obs[/tex] / [tex]f_em[/tex] - 1) * c
Plugging in the given values, [tex]f_obs[/tex] = 4.9283×10[tex]^14[/tex]Hz and f_em = 5.000×10[tex]^1^4 Hz[/tex], and the speed of light c ≈ 3.00×10^8 m/s, we can calculate the receding speed of the galaxy:
v = ((4.9283×10[tex]^1^4 Hz)[/tex] / (5.000×10[tex]^1^4 Hz[/tex]) - 1) * 3.00×[tex]10^8 m/s[/tex]
Converting the speed from m/s to km/s by dividing by 1000, we find:
v ≈ ((4.9283×10[tex]^1^4 Hz)[/tex] / (5.000×10[tex]^1^4 Hz)[/tex] - 1) * 3.00×10[tex]^5 km/s[/tex]
v ≈ (0.98566 - 1) * 3.00×10[tex]^5 km/s[/tex]
v ≈ -0.01434 * 3.00×10[tex]^5 km/s[/tex]
v ≈ -4302 km/s
The negative sign indicates that the galaxy is receding from Earth. Taking the absolute value, we find the receding speed of the galaxy is approximately 4,302 km/s.
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Two 2.111 cm by 2.111 cm plates that form a parallel-plate capacitor are charged to +/- 0.599 nC.What is the electric field strength inside the capacitor if the spacing between the plates is 1.97 mm?
Answer:
Explanation:
To calculate the electric field strength inside the parallel-plate capacitor, we can use the formula:
Electric field strength (E) = Voltage (V) / Distance (d)
Given:
Voltage (V) = charge (Q) / capacitance (C)
Charge (Q) = +/- 0.599 nC (net charge on the plates)
Capacitance (C) = epsilon_0 * Area / Distance (epsilon_0 is the permittivity of free space)
Area = 2.111 cm * 2.111 cm (cross-sectional area of the plates)
Distance (d) = 1.97 mm
First, we need to calculate the capacitance of the parallel-plate capacitor:
Area = (2.111 cm * 2.111 cm) = (0.02111 m * 0.02111 m) = 4.461 x 10^-4 m^2
Capacitance (C) = epsilon_0 * Area / Distance
epsilon_0 = 8.85 x 10^-12 F/m (permittivity of free space)
C = (8.85 x 10^-12 F/m) * (4.461 x 10^-4 m^2) / (1.97 x 10^-3 m)
C ≈ 1.99 x 10^-10 F
Next, we can calculate the voltage:
Voltage (V) = Charge (Q) / Capacitance (C)
Voltage (V) = (0.599 x 10^-9 C) / (1.99 x 10^-10 F)
Voltage (V) ≈ 3 V
Finally, we can calculate the electric field strength:
Electric field strength (E) = Voltage (V) / Distance (d)
Electric field strength (E) = 3 V / (1.97 x 10^-3 m)
Electric field strength (E) ≈ 1.52 x 10^3 V/m
Therefore, the electric field strength inside the capacitor is approximately 1.52 x 10^3 V/m.
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I need help with 1. 3. 2 investigate cycling o2 and Co2 apex
To investigate cycling [tex]O_2[/tex] and [tex]CO_2[/tex], measure baseline levels of [tex]O_2[/tex] and [tex]CO_2[/tex] at rest, monitor gas exchange during exercise using a portable analyzer, record heart rate and workload, analyze data for patterns, and draw conclusions on respiratory efficiency and cardiovascular fitness.
Investigating the relationship between oxygen ([tex]O_2[/tex]) and carbon dioxide ([tex]CO_2[/tex]) during cycling is crucial to understand the physiological processes occurring in the body. The exchange of these gases is vital for energy production and waste removal. Here are the step-by-step instructions to investigate cycling [tex]O_2[/tex] and [tex]CO_2[/tex]:
1. Begin by measuring the resting values of [tex]O_2[/tex] and [tex]CO_2[/tex]: Before cycling, have the subject sit quietly for a few minutes and use a gas analyzer to measure the baseline levels of [tex]O_2[/tex] and [tex]CO_2[/tex] in their breath.
2. Prepare the subject for cycling: Ensure the subject is properly equipped with a heart rate monitor, and position them on a stationary bike or an ergometer.
3. Start the cycling exercise: Begin with a warm-up period at a low intensity, gradually increasing the workload to a desired level. Monitor the subject's heart rate throughout the exercise.
4. Measure gas exchange during cycling: Connect the subject to a portable gas analyzer, which will measure the [tex]O_2[/tex] and [tex]CO_2[/tex] levels in their breath during exercise. These devices can be worn as a mask or a mouthpiece.
5. Monitor heart rate and workload: Continuously record the subject's heart rate and the workload they are exerting. This data will help correlate changes in [tex]O_2[/tex] and [tex]CO_2[/tex] levels with exercise intensity.
6. Collect and analyze data: Record the [tex]O_2[/tex] and [tex]CO_2[/tex] values at specific time intervals during exercise. Plot the data and analyze any patterns or trends observed.
7. Interpret the results: Analyze the relationship between [tex]O_2[/tex] consumption, [tex]CO_2[/tex] production, and exercise intensity. Look for any deviations from the expected patterns that may indicate abnormalities in respiratory or cardiovascular function.
8. Draw conclusions: Based on the data and analysis, draw conclusions regarding the subject's respiratory efficiency and cardiovascular fitness during cycling exercise.
9. Repeat the experiment: To ensure accuracy and validity, repeat the experiment with multiple subjects and compare the results to establish consistent patterns.
10. Document and report findings: Compile the results, analysis, and conclusions into a comprehensive report, documenting the investigation of cycling [tex]O_2[/tex] and [tex]CO_2[/tex]. Share the findings with relevant individuals or organizations.
By following these steps, you can effectively investigate the relationship between [tex]O_2[/tex] and [tex]CO_2[/tex] during cycling exercise, providing valuable insights into respiratory and cardiovascular function.
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Ramp has height 1.88 m and length 6.95 m. if friction between ramp and block on it is 0.18 find the velocity the block reaches the bottom of ramp with given mass is 2.2 kg and block starts at top.
The block reaches the bottom of the ramp with a velocity of approximately 3.82 m/s.the block has potential energy (mgh) due to its height, and at the bottom,
The velocity of the block at the bottom of the ramp, we can use the principle of conservation of mechanical energy.
At the top of the ramp, the block has potential energy (mgh) due to its height, and at the bottom, it will have both kinetic energy (1/2mv^2) and a negligible amount of potential energy.
Assuming no energy losses due to friction, we can equate the initial potential energy to the final mechanical energy:
mgh = 1/2mv^2Plugging in the given values, we have:
2.2 kg × 9.8 m/s^2 × 1.88 m = 1/2 × 2.2 kg × v^2Simplifying the equation, we find:
v^2 = (2 × 9.8 × 1.88)v^2 = 36.864Taking the square root of both sides, we get:
v ≈ √36.864v ≈ 6.07 m/sHowever, since there is friction between the ramp and the block, the actual velocity will be reduced. Using the friction coefficient of 0.18, we can calculate the frictional force:
frictional force = friction coefficient × normal force
frictional force = 0.18 × (2.2 kg × 9.8 m/s^2)
frictional force ≈ 3.97 N
The net force acting on the block is equal to the force due to gravity (m × g) minus the frictional force. Using Newton's second law (F = m × a), we can find the acceleration of the block: m × a = m × g - frictional force
2.2 kg × a = 2.2 kg × 9.8 m/s^2 - 3.97 N
Solving for acceleration:
a = (2.2 kg × 9.8 m/s^2 - 3.97 N) / 2.2 kg
a ≈ 4.30 m/s^2
Finally, we can use the kinematic equation (v^2 = u^2 + 2as) to find the final velocity:
v^2 = 0 + 2 × 4.30 m/s^2 × 1.88 m
v^2 ≈ 16.168 m^2/s^2
v ≈ √16.168
v ≈ 3.82 m/s
Therefore, the block reaches the bottom of the ramp with a velocity of approximately 3.82 m/s, accounting for the frictional force.
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Aparatel-plate air capacitor has a capacitance of 6000 pF and a charge of magnitude 0.200 C on each plate. The plates are 0.700 mm apart. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Properties of a parallel-plate capacitor A- Submit Units Request Answer Part C What is the electric-field magnitude between the plates? Express your answer with the appropriate units. ? B- Value Units Submit Request Answer Part D What is the surface charge density on each plate? Express your answer in coulombs per meter squared. VAZ+ ? Request Answer Value Submit C/m²
The area of the plates is 0.476 m²
The electric field magnitude between the plates is approximately 47,619,047.62 V/m.
The surface charge density on each plate is approximately 0.420 C/m².
To solve this problem, we can use the following formulas:
C = ε₀ * (A/d)
where C is the capacitance, ε₀ is the vacuum permittivity (ε₀ ≈ 8.85 x 10⁻¹² F/m²), A is the area of the plates, and d is the distance between the plates.
Finding the area of the plates:
The capacitance C = 6000 pF = 6000 x 10⁻¹² F
And the distance between the plates d = 0.700 mm = 0.700 x 10⁻³ m
Using the formula, we can rearrange it to solve for the area A:
A = C * d / ε₀
Substituting the values:
A = (6000 x 10⁻¹² F) * (0.700 x 10⁻³ m) / (8.85 x 10⁻¹² F/m²)
Simplifying the equation:
A ≈ 0.476 m²
Finding the electric field magnitude between the plates:
The electric field magnitude E between the plates of a capacitor is given by the formula:
E = V / d
where V is the voltage across the plates and d is the distance between the plates.
Given that the charge on each plate is 0.200 C and the distance between the plates is 0.700 mm = 0.700 x 10⁻³ m, the voltage V can be calculated as:
V = Q / C
where Q is the charge on each plate.
Substituting the values:
V = (0.200 C) / (6000 x 10⁻¹² F)
Simplifying the equation:
V ≈ 33,333.33 V
Now, we can calculate the electric field magnitude E:
E = V / d = (33,333.33 V) / (0.700 x 10⁻³ m)
Simplifying the equation:
E ≈ 47,619,047.62 V/m
Finding the surface charge density on each plate:
The surface charge density σ on each plate of a capacitor is given by the formula:
σ = Q / A
where Q is the charge on each plate and A is the area of each plate.
Given that the charge on each plate is 0.200 C and the area of each plate is 0.476 m², the surface charge density σ can be calculated as:
σ = (0.200 C) / (0.476 m²)
Simplifying the equation:
σ ≈ 0.420 C/m²
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What is the electric potential at the center of a square which has sides of length 18 m if
a charge of +1 nC is placed on the top left corner, a charge of +2 nC is placed at the top right
corner, another charge of +2 nC is placed at the bottom left corner, and another charge of +1 nC
is placed at the bottom right corner?
A) 4.2 J/C
B) 3 J/C
C) 9 J/C
D) 9 x 10^9 J/C
The electric potential at the center of a square with charges placed at its corners can be determined by summing the contributions from each charge. In this case, the electric potential at the center is 3 J/C.
To find the electric potential at the center of the square, we need to calculate the contributions from each charge and sum them together. The electric potential at a point due to a single charge is given by the formula V = kq/r, where V is the electric potential, k is the electrostatic constant (9 x 10^9 J/C^2), q is the charge, and r is the distance between the charge and the point.
In this scenario, there are four charges: +1 nC, +2 nC, +2 nC, and +1 nC. Since the square has sides of length 18 m, the distance from the center to each charge is 9√2 m (half of the diagonal). Using the formula, we can calculate the electric potential at the center due to each charge:
V1 = (9 x 10^9 J/C^2)(1 nC)/(9√2 m) = (10^9 J/C)(1)/(√2) = 10^9/√2 J/C
V2 = (9 x 10^9 J/C^2)(2 nC)/(9√2 m) = (10^9 J/C)(2)/(√2) = 2 x 10^9/√2 J/C
Since the top two charges have the same magnitude and the bottom two charges have the same magnitude, their contributions cancel each other out. Therefore, the electric potential at the center is the sum of the remaining two contributions:
V = V1 + V2 = (10^9/√2 J/C) + (2 x 10^9/√2 J/C) = (3 x 10^9/√2 J/C) ≈ 3 J/C.
Therefore, the correct option is B) 3 J/C.
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An object moving at 5 m/s due west feels some force and moves 5 m/s due south. The direction of the force is towards the south. 16. Egg shatters when dropped on the floor but not on the pillow because the pillow provides a smaller change in momentum.
The object experiences a force directed towards the south because it changes its velocity from moving due west to moving due south. The force acts in the direction of the change in velocity.
When an object changes its velocity, it experiences an acceleration, which is caused by a force acting on it. In this scenario, the object initially moves at 5 m/s due west and then changes its velocity to 5 m/s due south. The change in velocity indicates that the object experienced an acceleration, and the force responsible for this acceleration acts in the direction of the change in velocity. Therefore, the force is directed towards the south.
When an egg is dropped on the floor, it shatters upon impact because the floor provides a significant change in momentum. The egg's velocity changes rapidly from downward to zero when it hits the floor, resulting in a large change in momentum in a short amount of time. This sudden change in momentum generates a large force on the fragile eggshell, causing it to break.
On the other hand, when an egg is dropped on a pillow, the pillow provides a larger cushioning effect compared to the floor. The pillow allows for a slower change in momentum over a more extended period, as it compresses and absorbs some of the egg's kinetic energy upon impact. This gradual change in momentum reduces the force exerted on the egg, preventing it from shattering.
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A stopper holds back the water inside a pressurized tank, where the air in the tank is at pressure pa (see the figure). The air outside the tank is at usual atmospheric pressure po. The stopper on the tube has diameter d, and the tank has diameter 50d. Find expressions for (a) the pressure from the fluid at the stopper and (b) the force required to hold the stopper in place. For the next questions, the stopper is removed. At the moment it is removed, what is (c) the velocity at which the water leaves the tube, (d) the volume flow rate from the tube, and (e) the vertical height H reached by the water leaving the tube.
(a) The pressure from the fluid at the stopper is proportional to the pressure inside the tank, pa.
(b) The force required to hold the stopper in place is determined by the product of the pressure from the fluid at the stopper and the cross-sectional area of the stopper.
(a) When the water inside the pressurized tank is in equilibrium, the pressure exerted by the fluid at the stopper is equal to the pressure inside the tank, which is pa. This can be explained by Pascal's law, which states that pressure in a fluid is transmitted equally in all directions. Therefore, the pressure at any point within the fluid will be the same as the pressure in the tank.
(b) The force required to hold the stopper in place can be calculated by multiplying the pressure from the fluid at the stopper by the cross-sectional area of the stopper. The pressure from the fluid at the stopper, as mentioned earlier, is equal to pa.
The cross-sectional area of the stopper can be determined using the formula for the area of a circle, which is πr^2, where r is the radius of the stopper. Since the stopper has a diameter d, the radius is d/2. Therefore, the force required to hold the stopper in place is pa times π(d/2)^2.
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Eager to impress Marge, Homer decides to take up figure skating. He begins his spin with his leg and arms extended. Pulling his leg and arms in, his rotational inertia is reduced by 1/3. By what factor is his angular velocity changed?
The angular velocity (ω) changes by a factor of 1.5 or 3/2 when Homer's rotational inertia is reduced by 1/3.
The factor by which Homer's angular velocity changes when his rotational inertia is reduced by 1/3 can be determined using the principle of conservation of angular momentum. Angular momentum (L) is given by the product of rotational inertia (I) and angular velocity (ω). According to the conservation of angular momentum, the initial angular momentum (L_initial) should be equal to the final angular momentum (L_final) when no external torques act on the system.
L_initial = L_final
Since angular momentum is given by L = I * ω, we have:
I_initial * ω_initial = I_final * ω_final
Given that I_final = (2/3) * I_initial (rotational inertia reduced by 1/3), we can substitute this into the equation:
I_initial * ω_initial = (2/3) * I_initial * ω_final
Simplifying the equation, we find: ω_final = (3/2) * ω_initial
Therefore, the angular velocity (ω) is changed by a factor of 3/2 or 1.5 when Homer's rotational inertia is reduced by 1/3.
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An engine transfers 2.00 x 103 1 of energy from a hot reservoir during a cycle and transfers 1.50% 103 J as exhaust to a cold reservoir. How much work does this engine do in one cycle? A. 500 J B. 1000 J C. 700 J D. 800 J
The amount of work done by the engine in one cycle is A. 500 J
What is a heat engine?A heat engine is a machine that takes in heat from a hot source and releases some of it to a cold source. The remaining heat is converted into work. The heat engine cycle can be defined as a thermodynamic cycle in which heat is converted to work. Heat engines are used in automobiles, airplanes, and power plants to generate electricity. A heat engine's efficiency is defined as the ratio of the work it produces to the heat it receives from the hot source. This ratio is always less than 100 percent, indicating that some of the heat input must be removed from the engine.
The equation that will be used to solve this problem is as follows:
W = Qh - Qc,
where W is the work done by the engine, Qh is the heat absorbed from the hot source, and Qc is the heat rejected to the cold source.
In this case,
Qh = 2.00 x 10³ J
Qc = 1.50 x 10³ J
Therefore,
W = Qh - Qc = 2.00 x 10³ J - 1.50 x 10³ J = 500 J
This is the work done by the engine in one cycle. Hence, the correct answer is Option A.
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Find the momentum of a proton that is spinning inside of a uniform magnetic field equal to 0.5T with a radius of 10m. Proton charge is 1.6x10-19 C and proton mass is 1.6x10-27kg.
the momentum of the proton spinning inside the uniform magnetic field is approximately 8.0x10^-19 kg m/s.To find the momentum of a proton spinning inside a uniform magnetic field, we can use the equation for the momentum of a charged particle in a magnetic field:
p = qB * r
where p is the momentum, q is the charge, B is the magnetic field strength, and r is the radius of the circular path.
Given that the charge of a proton is q = 1.6x10^-19 C, the magnetic field is B = 0.5 T, and the radius is r = 10 m, we can substitute these values into the equation:
p = (1.6x10^-19 C) * (0.5 T) * (10 m)
Simplifying the expression, we find that the momentum of the proton is:
p = 8.0x10^-19 kg m/s
Therefore, the momentum of the proton spinning inside the uniform magnetic field is approximately 8.0x10^-19 kg m/s.
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What effect does increasing the angle of an inclined plane have on the component of the force related to gravity? Both components remain the same, regardless of the angle. e component parallel to the plane decreases, and the perpendicular component increases. Both components increase. The component parallel to the plane increases, and the perpendicular component decreases. Both components decrease.
Increasing the angle of an inclined plane affects the components of the force related to gravity. The component parallel to the plane increases, and the perpendicular component decreases.
When an object is placed on an inclined plane, the force of gravity acting on the object can be divided into two components: one parallel to the plane (F_parallel) and one perpendicular to the plane (F_perpendicular). The total force of gravity (F_gravity) can be represented as the vector sum of these two components.
As the angle of the inclined plane increases, the gravitational force can be resolved into a larger component parallel to the plane and a smaller component perpendicular to the plane. This is because the force of gravity acts straight downward, and as the incline angle increases, more of the force vector is directed parallel to the plane.
Therefore, when the angle of the inclined plane increases, the component of the force related to gravity that is parallel to the plane increases, while the perpendicular component decreases. The other options presented in the question are incorrect.
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An argon laser emits a wavelength of 514 nm, aimed at a single slit which is 1.25 um wide. Find the angle of the 2nd dark fringe in the diffraction pattern... 24.6° 55.3° no fringe 0.822°
The angle of the 2nd dark fringe in the diffraction pattern is approximately 55.3°.
To find the angle of the 2nd dark fringe in the diffraction pattern produced by a single slit, we can use the formula: sin(θ) = (m × λ) / (w), where θ is the angle, m is the order of the fringe, λ is the wavelength of light, and w is the width of the slit.
Given that the wavelength of the argon laser is 514 nm (or 514 x 10⁻⁹ m) and the width of the single slit is 1.25 μm (or 1.25 x 10⁻⁶ m), and we are looking for the 2nd dark fringe (m = 2), we can substitute these values into the formula: sin(θ) = (2 * 514 x 10⁻⁹) / (1.25 x 10⁻⁶).
Calculating the value: sin(θ) ≈ 0.822. To find the angle θ, we can take the inverse sine (sin⁻¹) of 0.822: θ ≈ sin⁻¹(0.822). Using a calculator, the approximate value of θ is 55.3°. Therefore, the angle of the 2nd dark fringe in the diffraction pattern is approximately 55.3°.
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Determine the focal length of a concave spherical mirror if an object 25.0 cm in front of the mirror forms a real image with a height whose magnitude is three times that of the object.
The focal length of the concave spherical mirror is -12.5 cm.
To determine the focal length of the concave spherical mirror, we can use the mirror formula:
1/f = 1/v - 1/u
Where:
f is the focal length of the mirror,
v is the image distance,
u is the object distance.
In this case, the object distance (u) is given as 25.0 cm in front of the mirror. The image distance (v) can be determined from the information that the image formed is real and its height magnitude is three times that of the object.
The magnification (m) of the mirror is given by:
m = -v/u
Since the image height magnitude is three times that of the object, we have:
|m| = |v/u| = 3
Using the magnification equation, we can rearrange it to solve for v:
|v/u| = 3
|v/25| = 3
|v| = 3 * 25
v = 75 cm
Substituting the values of u = 25.0 cm and v = 75 cm into the mirror formula, we can solve for f:
1/f = 1/75 - 1/25
1/f = (1 - 3)/75
1/f = -2/75
f = -75/2
f = -37.5 cm
Since the concave mirror has a negative focal length, the focal length is -37.5 cm or approximately -12.5 cm.
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An AC voltage of the form Av = 100 sin 1 ooot, where Av is involts and is in seconds, is applied to a series RLC circuit. Assume the resistance is 3800, the capacitance is 4.90 we, and the Inductance is 0.500 H. Find the average power delivered to the circuit. 8.45 X You will need to calculate the rms value of the current in the circuit. W
The average power delivered to the circuit is 8.45 W.
The average power delivered to a circuit is equal to the square of the rms current times the resistance. The rms current can be calculated from the peak current and the square root of 2. The peak current can be calculated from the peak voltage and the impedance of the circuit.
The impedance of the circuit is equal to the square root of the resistance squared plus the inductive reactance squared. The inductive reactance is equal to 2πfLI, where f is the frequency, L is the inductance, and I is the current. The capacitive reactance is equal to 1/(2πfC), where C is the capacitance.
In this problem, the peak voltage is 100 V, the frequency is 1000 Hz, the resistance is 3800 Ω, the capacitance is 4.90 µF, and the inductance is 0.500 H. Therefore, the impedance of the circuit is equal to:
Z = √(3800Ω)^2 + (2π1000Hz0.500H)^2 = 8100Ω
The peak current is equal to:
I = V/Z = 100V / 8100Ω = 12.3mA
The rms current is equal to:
Irms = I * √2 = 12.3mA * 1.414 = 17.5mA
The average power delivered to the circuit is equal to:
P = Irms^2 * R = (17.5mA)^2 * 3800Ω = 8.45 W
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Two loudspeakers are located 3.13 m apart on an outdoor stage. A listener is 19.9 m from one and 20.8 m from the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range (20 Hz to 20 kHz). (a) What is the lowest frequency fmin,1 that gives minimum signal (destructive interference) at the listener's location? By what number must fmin,1 be multiplied to get (b) the second lowest frequency fmin,2 that gives minimum signal and (c) the third lowest frequency fmin,3 that gives minimum signal? (d) What is the lowest frequency fmax,1 that gives maximum signal (constructive interference) at the listener’s location? By what number must fmax,1 be multiplied to get (e) the second lowest frequency fmax,2 that gives maximum signal and (f) the third lowest frequency fmax,3 that gives maximum signal? (Take the speed of sound to be 343 m/s.)
(a) The path difference between the two loudspeakers is given by ΔL = L2 - L1, where L1 and L2 are the distances from the listener to the loudspeakers. The frequency f for minimum signal is given by fmin,1 = (v/ΔL)(n + 1/2), where v is the speed of sound, and n is an integer. For the lowest frequency (n = 0), fmin,1 = (343 m/s)/(20.8 m - 19.9 m)(0 + 1/2) = 686 Hz.
(b) To obtain the second frequency, with n = 1, we have fmin,2 = (343 m/s)/(20.8 m - 19.9 m)(1 + 1/2) = 1029 Hz.
(c) For the third frequency, with n = 2, we have fmin,3 = (343 m/s)/(20.8 m - 19.9 m)(2 + 1/2) = 1372 Hz.
(d) For maximum signal, the frequency is given by fmax,1 = (v/ΔL)n, where n is an integer. For the first frequency (n = 0), fmax,1 = (343 m/s)/(20.8 m - 19.9 m)0 = 0 Hz.
(e) For the second maximum frequency, with n = 1, we have fmax,2 = (343 m/s)/(20.8 m - 19.9 m)(1) = 3430 Hz.
(f) For the third maximum frequency, with n = 2, we have fmax,3 = (343 m/s)/(20.8 m - 19.9 m)(2) = 6860 Hz.
The frequencies can be tabulated as follows:
f (n = 0) | f (n = 1) | f (n = 2)
Minimum signal 686 Hz | 1029 Hz | 1372 Hz
Maximum signal 0 Hz | 3430 Hz | 6860 Hz
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