The force exerted by the water on the bottom of the pool is 4,269,312 N and on each end of the pool is 227,711 N.
What is the force exerted by the water on the bottom and each end of a swimming pool?(a) The force exerted by the water on the bottom of the pool is equal to the weight of the water above it.
The weight of the water is equal to its mass multiplied by the acceleration due to gravity (g = 9.81 m/s^2), and its mass is equal to its volume multiplied by its density (ρ = 1000 kg/m^3 for fresh water at standard conditions):
[tex]Volume\ of\ water = length * width * depth = 32.0 m * 8.0 m * 1.70 m = 435.2 m^3\\Mass of water = Volume x Density = 435.2 m^3 * 1000 kg/m^3 = 435200 kg\\Weight of water = Mass x g = 435200 kg x 9.81 m/s^2 = 4,269,312 N[/tex]
Therefore, the force exerted by the water on the bottom of the pool is 4,269,312 N.
(b) The force exerted by the water on each end of the pool can be calculated by considering the pressure of the water on the end. The pressure of the water at a depth of 1.70 m is equal to the height of the water column (1.70 m) multiplied by the density of water and the acceleration due to gravity:
[tex]Pressure = depth * density * g = 1.70 m * 1000 kg/m^3 * 9.81 m/s^2 = 16,743 Pa[/tex]
The force exerted by the water on each end is equal to the pressure multiplied by the area of the end:
[tex]Force = Pressure * Area = 16,743 Pa * 8.0 m * 1.70 m = 227,711 N[/tex]
Therefore, the force exerted by the water on each end of the pool is 227,711 N.
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The waters off the coast of Iceland are filled with pods of killer whales, which migrate there during the summer. Wildlife parks that rely on the killer whales for entertainment hunt the killer whale almost exclusively in the water of Iceland, because strict sanctions forbid them from doing so off the coast of North America, an area also abundant in killer whales. Since Iceland recently gave in to pressure from international groups opposed to the hunting of killer whales, it too will forbid the hunting of killer whales off its coast. Therefore, all wildlife parks will be forced to end their shows featuring killer whales once their current killer whales are unable to perform. " All of the following cast doubt on the conclusion of the argument EXCEPT?
The fact that Iceland recently gave in to pressure from international groups opposed to the hunting of killer whales and will forbid the hunting of killer whales off its coast does not cast doubt on the conclusion of the argument.
It actually supports the conclusion that all wildlife parks will be forced to end their shows featuring killer whales once their current killer whales are unable to perform.
The argument states that wildlife parks rely on hunting killer whales exclusively in the waters of Iceland because they are forbidden to do so off the coast of North America. If Iceland also forbids hunting, the parks will no longer have a source of killer whales, which will ultimately lead to the end of their shows. The conclusion is directly supported by the premise, and there is noccontradictory information provided that would cast doubt on this logical chain of reasoning.
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You put a 51.7 gram mass on a spring, set it in motion with a small amplitude, and count 21 cycles. Those 21 cycles took 3.42 seconds What is kHM Answer
The spring constant for a mass of 51.7 grams on a spring that undergoes 21 cycles with a small amplitude in 3.42 seconds is 76.8 N/m.
The value of k for a mass on a spring can be determined using the formula T=2π√(m/k), where T is the period of oscillation, m is the mass, and k is the spring constant. In this problem, we know that the mass is 51.7 grams and that 21 cycles took 3.42 seconds, which means that the period of oscillation is T=3.42/21=0.163 seconds. Since the amplitude is small, we can assume that the motion is simple harmonic, which means that T=2π√(m/k) can be used. Rearranging this formula gives k=m(2π/T)^2, which gives k=51.7(2π/0.163)^2=76.8 N/m.
This value was calculated using the formula k=m(2π/T)^2, where m is the mass and T is the period of oscillation.
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A charge q1 = 2 µc is at the origin, and a charge q2 = 10 µc is on the x axis at x = 10 m. find the force on charge q2 . the colulomb constant is 8.98755 × 109 n · m 2 /c 2 . answer in units of n.
The force on charge q2 is approximately 179.751 N.
The force between two point charges can be found using Coulomb's law:
F = (k * q1 * q2) / r^2
Where F is the force between the charges, k is the Coulomb constant (8.98755 × 10^9 N·m^2/C^2), q1 and q2 are the magnitudes of the charges in Coulombs, and r is the distance between the charges in meters.
In this case, q1 = 2 µC and q2 = 10 µC. The distance between the charges is the distance between the origin and the point on the x-axis where q2 is located, which is 10 m.
So, we can calculate the force on q2 as follows:
F = (8.98755 × 10^9 N·m^2/C^2) * (2 µC) * (10 µC) / (10 m)^2
F = (8.98755 × 10^9 * 2 * 10) / 100
F = 1.79751 × 10^9 / 100
F = 1.79751 × 10^7 N
The force on charge q2, we can use Coulomb's Law. Coulomb's Law states that the force (F) between two point charges is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (r) between them:
F = k * (q1 * q2) / r^2
In this case, q1 = 2 µC, q2 = 10 µC, r = 10 m, and the Coulomb constant (k) is 8.98755 × 10^9 N·m^2/C^2.
The charges to Coulombs: q1 = 2 × 10^-6 C and q2 = 10 × 10^-6 C.
F = (8.98755 × 10^9 N·m^2/C^2) * ((2 × 10^-6 C) * (10 × 10^-6 C)) / (10 m)^2
F = (8.98755 × 10^9 N·m^2/C^2) * (2 × 10^-5 C^2) / (100 m^2)
F = 179.751 N
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If a person goes to the bottom of a very deep mine shaft on a planet of uniform density, which of the following is true? 2. (A) The person's weight is exactly the same as at the surface. (B) The person's weight is less than at the surface. (C) The person's weight is greater than at the surface. (D) The person's weight may increase or decrease, depending on the density of the planet.
If a person goes to the bottom of a very deep mine shaft on a planet of uniform density, then the person's weight is exactly the same as at the surface. Option(A) is true.
The force of gravity is directly proportional to the mass of the planet and inversely proportional to the square of the distance between the person and the center of the planet.
Gravity is a fundamental force that governs the motion of objects in the universe. It is an attractive force between any two objects with mass or energy, and its strength depends on the mass and distance between the objects.
Since the planet has uniform density, the mass beneath the person cancels out, resulting in no change in weight.
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The dc source supplying an inverter with a bipolar PWM output is 96 V. The load is an RL
series combination with R =32 ohms and L = 24 mH. The output has a fundamental frequency of 60 Hz.
a) Specify the amplitude modulation ratio to provide a 54-V rms fundamental frequency output.
b) If the frequency modulation ratio is 17, determine the total harmonic distortion of the load current.
c) What is the real power absorbed by the load?
a) The amplitude modulation ratio is 0.5625.
b) The total harmonic distortion of the load current is 40.53%.
c) The real power absorbed by the load is 405.36 W.
a) The amplitude modulation ratio (m) can be calculated using the equation m = Erms/Emax, where Erms is the desired rms voltage and Emax is the maximum voltage. Here, Erms is given as 54 V and Emax can be calculated as Emax = √(2) × 96 V = 135.76 V. Substituting these values, we get m = 0.5625.
b) The frequency modulation ratio (f) is given as 17. The total harmonic distortion (THD) can be calculated using the equation THD = √((Vthd² - V1²)/V1²) * 100%, where Vthd is the total harmonic voltage and V1 is the fundamental voltage. Here, V1 is given as 54 V. The total harmonic voltage can be calculated as Vthd = sqrt(sum of squares of all harmonic voltages) = sqrt((V3² + V5² + V7² + ...)/2), where V3, V5, V7, etc., are the harmonic voltages. Substituting the given values, we get Vthd = 43.72 V. Substituting these values, we get THD = 40.53%.
c) The real power absorbed by the load can be calculated using the equation P = V1²/R, where V1 is the fundamental voltage and R is the resistance of the load. Substituting the given values, we get P = 405.36 W.
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A square-wave inverter supplies an RL series load with R=25 ohms and L=25mH. The output frequency is 120 Hz. (a) Specify the dc source voltage such that the load current at the fundamental frequency is 2.0 A rms. (b) Determine the THD of the load current (until 9), show all your work. + Vdc
(a) The dc source voltage is 61.2 V.
(b) The THD of the load current is approximately 33.2%.
(a) To calculate the dc source voltage required to produce a load current of 2.0 A rms, we first need to calculate the impedance of the load at the fundamental frequency. The impedance can be calculated as Z = R + jωL, where R is the resistance of the load, L is the inductance of the load, and ω is the angular frequency.
ω = 2πf
ω = 2π x 120 Hz
ω = 753.98 rad/s
Z = 25 + j(753.98 x 0.025)
Z = 25 + j18.85 Ω
The rms value of the load current is given by I = V/Z, where V is the rms value of the voltage supplied by the inverter.
I = 2.0 A rms, Z = 25 + j18.85 Ω
Therefore, V = IZ
V = (2.0 A rms) x (25 + j18.85 Ω)
V = 61.2 + j45.35 V rms
The dc source voltage is the average value of the voltage waveform, which is equal to the rms value multiplied by π/2.
Vdc = (π/2) x 61.2 V rms ≈ 96.2 Vdc
(b) The total harmonic distortion (THD) of the load current is a measure of the distortion of the current waveform from a perfect sinusoid. It is defined as the square root of the sum of the squares of the harmonic components of the current waveform, divided by the rms value of the fundamental component.
THD = √[(I2² + I3² + ... + In²)/I1²] x 100%
where I1 is the rms value of the fundamental component, and I2, I3, ..., In are the rms values of the second, third, ..., nth harmonic components.
For a square-wave inverter, the load current waveform contains only odd harmonic components. The rms value of the nth harmonic component can be calculated as
In = (4Vdc/(nπZ)) x sin(nπ/2)
where n is the harmonic number.
Using this equation, we can calculate the rms values of the first three harmonic components of the load current.
I1 = 2.0 A rms (given)
I3 = (4 x 96.2 Vdc / (3π x 25 Ω)) x sin(3π/2)
I3 ≈ 0.632 A rms
I5 = (4 x 96.2 Vdc / (5π x 25 Ω)) x sin(5π/2)
I5 ≈ 0.254 A rms
The THD can now be calculated as
THD = √[(0.632² + 0.254²)/2.0²] x 100%
THD ≈ 33.2%
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Consider the de Broglie wavelength of an electron What is the de Broglie wavelength of an electron traveling at a speed of 5.0×106 m/s? Give your answer in pm ト Grade Summary Deductions Potential pm 0% 100% Submissions tan() | π | ( 789 cosO cotanO asin0 acos0 atan acotan0 sinh coshO tanh0 cotanh0 °Degrees -Radians sin Attempts remaining: 999 % per attempt) detailed view 0 END vo DELCLEAR Submit I give up! Hints: for a .0%-deduction. Hints remaining: 0 Feedback: 5%-deduction per feedback.
The de Broglie wavelength of an electron traveling at a speed of 5.0 x 10^6 m/s is approximately 0.145 picometers (pm).
What is the equation for calculating the de Broglie wavelength of an electron, and what is the de Broglie wavelength of an electron traveling at a speed of 5.0 x 10^6 m/s?The de Broglie wavelength of an electron is given by the equation:
λ = h/mv
Where λ is the de Broglie wavelength, h is Planck's constant, m is the mass of the electron, and v is the velocity of the electron.
Substituting the given values, we get:
λ = h/(mv) = (6.626 x 10^-34 J s)/(9.11 x 10^-31 kg x 5.0 x 10^6 m/s)
λ = 0.145 pm (rounded to three significant figures)
Therefore, the de Broglie wavelength of an electron traveling at a speed of 5.0 x 10^6 m/s is approximately 0.145 picometers (pm).
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david bowie changed his original name to avoid confusion with which famous dave?
David Bowie changed his original name (David Robert Jones) to avoid confusion with Davy Jones, a member of the popular band The Monkees.
Bowie didn't want to be associated with Davy Jones and sought a distinct identity for his own career in music. Davy Jones was a British singer and actor who gained fame as a member of The Monkees in the 1960s. As David Robert Jones began his own musical journey, he decided to adopt the stage name "David Bowie" to prevent any potential confusion between the two artists. Bowie's new name not only provided him with a unique identity but also allowed him to craft a distinct image and persona that would define his groundbreaking and influential career in music and art.
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Oblem 26. 53 - Enhanced - with Feedback
You have a semicircular disk of glass with an index of
ofraction of n = 156 (Figure 1) You may want to review
Pages 929 - 936)
Part A
Find the incident angle o for which the beam of light in the figure will hit the indicated point on the screen
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The incident angle θ for which the beam of light in the figure will hit the indicated point on the screen is 60 degrees.
In this question, we need to find the incident angle for which the beam of light in the figure will hit the indicated point on the screen. We have a semicircular disk of glass with an index of fraction of n = 156 (Figure 1). We are given that the refractive index of the glass is n = 1.56. Using Snell's law,n1sinθ1=n2sinθ2where, n1= refractive index of the incident medium, n2= refractive index of the refracted medium, θ1= angle of incidence, θ2= angle of refraction. As air is the incident medium, the refractive index of air is 1.n1 = 1 and n2 = 1.56 sin(θ1) = 1.56sin(θ2)
As the angle of incidence (i) and the angle of reflection (r) are equal,i = rso, the angle between the incident ray and the normal, θ1 = 60°
Thus, sin(60) = 1.56sin(θ2)sin(θ2) = 0.63θ2 = 40.94°
As the light is refracted away from the normal, the angle of incidence is greater than the angle of refraction.
Hence, the incident angle of the beam of light is 60°.
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