The energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed is 6.34 J and the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed is 11.20 J.
Energy stored in vacuum capacitor (U₁) = 6.34 JInitial separation between the plates (d₁) = 3.90 mm
Final separation between the plates (d₂) = 1.50 mm
Part A: If the capacitor was disconnected from the potential source before the separation of the plates was changed, then the energy stored will remain constant as the charge stored in the capacitor will not change.
Thus, Energy stored in the capacitor after changing the separation of the plates = 6.34 J.
Part B: If the capacitor remained connected to the potential source while the separation of the plates was changed, then the charge stored in the capacitor will increase as the capacitance of the capacitor is inversely proportional to the distance between the plates
i.e., as the separation decreases the capacitance increases.
The formula to find the capacitance of the capacitor is given by,C = ε₀A/d
Where C is the capacitance, A is the area of each plate, d is the separation between the plates, and ε₀ is the permittivity of free space.
The energy stored in the capacitor can be given as,U = 1/2 CV²where V is the potential difference between the plates
Substituting the value of C in the above equation, we get:U = (ε₀A/2d) V²As the capacitor remains connected to the potential source, the potential difference between the plates will also remain constant and equal to the potential difference provided by the potential source.
Now, the capacitance after changing the separation of the plates can be calculated as:C' = ε₀A/d₂
Substituting the values of A, d₁ and d₂ in the above equation, we get:C' = 8.854 x 10⁻¹² x 0.003²/0.0015C' = 3.542 x 10⁻¹⁰ F
The energy stored in the capacitor after changing the separation of the plates can be calculated as:U' = (ε₀A/2d₂) V²Substituting the values of A, d₂ and V in the above equation,
we get:U' = (8.854 x 10⁻¹² x 0.003²/2 x 0.0015) (V)²U' = 1.77 (V)²
Therefore, the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed is 1.77 times the initial energy stored i.e.,U' = 1.77 x 6.34U' = 11.20 J.
Hence, the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed is 6.34 J and the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed is 11.20 J.
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300 g of water is brought to boiling temperature. The water is then left to cool to room temperature (25°C). The specific heat heat capacity is 4200 J/kg°C. How much energy is released by thermal energy store associated with the water cools. Show working
Answer:
94.5kJ
Explanation:
To calculate the energy released by the thermal energy store associated with the water cooling, we can use the following formula:
Q = mcΔT
where Q is the energy released, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
We first need to calculate the temperature change of the water. The initial temperature of the water is the boiling point of 100°C, and the final temperature is the room temperature of 25°C. Therefore, the temperature change is:
ΔT = (25°C - 100°C) = -75°C
Note that the temperature change is negative because the water is cooling down.
Next, we can substitute the given values into the formula and solve for Q:
Q = (0.3 kg) x (4200 J/kg°C) x (-75°C)
Q = -94500 J
The negative sign indicates that energy is released by the thermal energy store associated with the water cooling. Therefore, the energy released is 94,500 J, or approximately 94.5 kJ.
During 9.839.83 s, a motorcyclist changes his velocity from
1,x=−41.1v1,x=−41.1 m/s and 1,y=14.7v1,y=14.7 m/s to
2,x=−23.7v2,x=−23.7 m/s and 2,y=28.9v2,y=28.9 m/s.
During 9.839.83 s, a motorcyclist changes his velocity from 1,x=−41.1v1,x=−41.1 m/s and 1,y=14.7v1,y=14.7 m/s to 2,x=−23.7v2,x=−23.7 m/s and 2,y=28.9v2,y=28.9 m/s.
During the time interval of 9.83 s, a motorcyclist's velocity changes from (-41.1 m/s, 14.7 m/s) to (-23.7 m/s, 28.9 m/s). The initial velocity of the motorcyclist (v1) is (-41.1 m/s, 14.7 m/s).
The final velocity of the motorcyclist (v2) is (-23.7 m/s, 28.9 m/s).
The magnitude of the change in velocity (|Δv|) can be calculated using the formula:
|Δv| = √[(v2,x - v1,x)² + (v2,y - v1,y)²]
|Δv| = √[(-23.7 - (-41.1))² + (28.9 - 14.7)²]
|Δv|= √[322.56 + 202.5]
|Δv| = √525.06
|Δv| = 22.92 m/s
The direction of the change in velocity (θ) can be calculated using the formula:
θ = tan⁻¹[(v2,y - v1,y) / (v2,x - v1,x)]
θ = tan⁻¹[(28.9 - 14.7) / (-23.7 - (-41.1))]
θ = tan⁻¹[14.2 / 17.4]
θ = 42.1°
The change in velocity is 22.92 m/s in the direction of 42.1°.
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3cos(wt/3), where I is in meters and t in seconds. The acceleration of the particle The position of a partide is given by in my when2mis 249 2.1 275 228
The given equation represents the position of a particle as a function of time, given by x(t) = 3cos(wt/3), where x is in meters and t is in seconds. To find the acceleration of the particle, we need to take the second derivative of the position function with respect to time.
The first derivative of x(t) gives us the velocity function v(t):
v(t) = dx(t)/dt = -3w/3 * sin(wt/3)
Differentiating again, we find the second derivative, which is the acceleration function a(t):
a(t) = dv(t)/dt = d²x(t)/dt² = (-3w/3)² * cos(wt/3)
Simplifying further, we get:
a(t) = w² * cos(wt/3)
The acceleration of the particle, a(t), is given by w² times the cosine of wt/3.
In the given context, the values of w, which is the angular frequency, are not provided. Therefore, we cannot determine the specific numerical value of the acceleration. However, we can analyze its behavior based on the equation. The acceleration is directly proportional to w², meaning that increasing the value of w will result in a larger acceleration. Additionally, the cosine function oscillates between -1 and 1, so the acceleration will oscillate between -w² and w².
In summary, the acceleration of the particle is given by the equation a(t) = w² * cos(wt/3). The specific numerical value of the acceleration depends on the value of the angular frequency w, which is not provided in the given information.
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Look at the circuit diagram.
What type of circuit is shown?
open series circuit
open parallel circuit
closed series circuit
closed parallel circuit
The type of circuit shown in the diagram is a closed series circuit. The Option C.
What type of circuit is depicted in the circuit diagram?The circuit diagram illustrates a closed series circuit, where the components are connected in a series, forming a single loop. In a closed series circuit, the current flows through each component in sequence, meaning that the current passing through one component is the same as the current passing through the other components.
The flow of current is uninterrupted since the circuit forms a complete loop with no breaks or open paths. Therefore, the correct answer is a closed series circuit.
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How wide is the central maximum in degrees and cm? (wavelength=670nm) (L=30.0cm) (w=1.2E-5m)
To calculate the width of the central maximum in degrees, we can use the formula: θ = λ / w
The width of the central maximum is approximately 1.6749 cm.
The width of the central maximum is approximately 3.19 degrees.
Given:
Wavelength (λ) = 670 nm = 670 × 10⁻⁹ m
Width of the slit (w) = 1.2 × 10⁻⁵ m
Substituting these values into the formula:
θ = (670 × 10⁻⁹ m) / (1.2 × 10⁻⁵ m)
θ ≈ 0.05583 radians
To convert the angular width from radians to degrees, we can use the conversion factor:
1 radian = 180 degrees / π
θ° = θ × (180 degrees / π)
θ° ≈ 3.19 degrees
Therefore, the width of the central maximum is approximately 3.19 degrees.
To calculate the width of the central maximum in centimeters, we can use the formula:
Width(cm) = L × θ
where L is the distance from the slit to the screen and θ is the angular width.
Given:
Distance from the slit to the screen (L) = 30.0 cm
Substituting the values:
Width(cm) = (30.0 cm) × (0.05583 radians)
Width(cm) ≈ 1.6749 cm
Therefore, the width of the central maximum is approximately 1.6749 cm.
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Explain why the Sun appears to move through the stars during the course of a year. How does the Sun's motion through the stars affect the constellations seen in the nighttime sky? 1. How is the distribution of electrons amone the perabiele ererzs levels in a degenerate cas diflerent than that in an ordinary gas? Mow do the properties of a degenerate tat satter from those of an ordinary gas? 2. How do astronomers know that the formation of planetary nebulae is a common occurtence dutime the evolution of medium-mass stars? B 3. Why do the stars in a cluster evolve at different rates? Explain how the H-R diagram of a cluster of stars can be used to find the age of the cluster. 4. Explain how the distance to a Cepheid variable star can be determined from its light curve.
The relationship between a Cepheid variable's luminosity and pulsation period has been established as a way to estimate the distance to the star.
How is the distribution of electrons among the probable energy levels in a degenerate case different from that in an ordinary gas? How do the properties of a degenerate gas differ from those of an ordinary gas? In a degenerate gas, the electrons are compacted in the lower energy levels and become tightly jammed. As a result, their distribution varies from the probable energy levels predicted by the Maxwell-Boltzmann statistics. The most important property of a degenerate gas is that its pressure is not connected to its temperature, unlike an ordinary gas. When the pressure of an ordinary gas is decreased, the molecules move slower, and the temperature drops. This is not the case with a degenerate gas. Because of the limitations of quantum mechanics, the electrons in a degenerate gas are so tightly packed that they cannot be further compressed. The gas pressure is caused by electron compression and is proportional to the number of electrons in the gas.
How do astronomers know that the formation of planetary nebulae is a common occurrence during the evolution of medium-mass stars? Astronomers know that planetary nebulae formation is a common event during the evolution of medium-mass stars since roughly 10% of all stars have a mass between 1 and 8 solar masses. These stars lose a large portion of their original mass when they transform into planetary nebulae in the later phases of their lives. Planetary nebulae may have played a crucial role in the formation of the Milky Way's interstellar medium and the cycles of star formation and interstellar matter redistribution that exist in the universe.
Why do the stars in a cluster evolve at different rates? Explain how the H-R diagram of a cluster of stars can be used to find the age of the cluster. The stars in a cluster evolve at different rates due to variations in their initial mass. Massive stars, for example, evolve much more quickly than less massive stars and die as supernovae. Star clusters are valuable laboratories for testing our theories about stellar evolution since all of the stars were formed at the same time from the same material. By analyzing the H-R diagram of a star cluster, we can determine the age of the cluster. This is due to the fact that the brightness and surface temperature of a star are both dependent on its mass and stage of evolution.
Explain how the distance to a Cepheid variable star can be determined from its light curve. The relationship between a Cepheid variable's luminosity and pulsation period has been established as a way to estimate the distance to the star. The period of a Cepheid variable star is directly linked to its absolute luminosity: brighter stars have longer periods. When we determine the star's period and apparent brightness, we can use this relationship to calculate the star's absolute brightness. The distance to the star may be calculated once we know its actual brightness and apparent brightness. The period-luminosity relationship for Cepheid variables was discovered by Henrietta Swan Leavitt in 1912.
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Find the magnitude and the direction of the magnetic field that will cause the electron to cross x=42 cme magnitude direction (b) What work is done on the electron during this motion? (c) How long will the trip take from y-axis to x-axis?
a)the magnitude of the magnetic field.B = 3.53 x 10^(-3) T and the magnetic field is directed in the negative z-direction.b)Work done by the magnetic field is zero because the magnetic field is perpendicular to the direction of motion.c) the time taken.t = 7.43 x 10^(-8) s.
A magnetic field that will cause the electron to cross x = 42 cm is given by (a) and (b). What work is done on the electron during this motion and how long will the trip take from the y-axis to the x-axis? Find the magnitude and direction of the magnetic field.Answer:Magnitude of magnetic field = 3.53 x 10^(-3) TDirection of magnetic field = Inverted in z-direction.
Work done = 0JTime taken = 7.43 x 10^(-8) sStep-by-step
A force exists on a charged particle due to the magnetic field, which results in circular motion. The strength of the magnetic force is given by the equation Fm = qvBsinθ, where q is the charge on the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
Lorentz force is the result of the magnetic force acting on a charged particle in a magnetic field, which causes the particle to move in a circle, as shown below:Fm = q(v×B)Here, B is the magnetic field vector, which is perpendicular to the plane of the paper. As a result, the force on the particle is perpendicular to its velocity vector and is directed towards the center of the circle.Force = maSo, ma = q(v×B)From this we get acceleration of the charged particle due to magnetic field.
By using this acceleration we can calculate the radius of the circle that the electron moves. As the path of electron is circular, centripetal force must be equal to the magnetic force.Fc = FmBy using these we can calculate the magnetic field magnitude, direction and work done and time taken.
(a) Magnitude and direction of the magnetic fieldAs the magnetic force is the centripetal force we haveFc = FmFrom this we getqvB = mv^2 / rB = mv / qr = mv / qBvSubstitute the values givenm = 9.11 x 10^(-31)kgq = 1.60 x 10^(-19) C x = 42 cm = 0.42 mT = 2.35 x 10^(-6) sB = m * v / (q * r)Calculate the magnitude of the magnetic field.B = 3.53 x 10^(-3) T
We know that the force is perpendicular to the velocity and the direction of the magnetic field is given by the right-hand rule. In the z-direction, the velocity vector is towards the observer, and the magnetic force vector is in the opposite direction to the observer. As a result, the magnetic field is directed in the negative z-direction.
(b) Work done by the magnetic field is zero because the magnetic field is perpendicular to the direction of motion. The magnetic field only causes a change in direction.
(c) As the magnetic force is the centripetal force we haveqvB = mv^2 / rBy substituting the valuesq = 1.60 x 10^(-19) Cv = 3.0 x 10^6 m/sm = 9.11 x 10^(-31) kgB = 3.53 x 10^(-3) Tr = 0.42 m Calculate the time taken.t = 7.43 x 10^(-8) s
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An AC generator supplies an rms voltage of 115 V at 60.0 Hz. It is connected in series with a 0.200 H inductor, a 4.60 uF capacitor and a 336 2 resistor. What is the impedance of the circuit? What is the rms current through the resistor?
What is the average power dissipated in the circuit?
What is the peak current through the resistor?
What is the peak voltage across the inductor?
What is the peak voltage across the capacitor? The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency?
The impedance of the circuit is 336.2 ohms. The rms current through the resistor is 0.342 A. The average power dissipated in the circuit is 39.2 W. The peak current through the resistor is 0.484 A. The peak voltage across the inductor is 68.7 V. The peak voltage across the capacitor is 19.6 V. The new resonance frequency is 60.0 Hz.
To find the impedance of the circuit, we need to consider the combined effects of the inductor, capacitor, and resistor. The impedance of an RL circuit is given by Z = [tex]\sqrt{(R^2 + (ωL - 1/(ωC))^2)}[/tex], where R is the resistance, ω is the angular frequency (2πf), L is the inductance, and C is the capacitance. Plugging in the values, we get Z = [tex]\sqrt{(336^2 + (2\pi (60)(0.200) - 1/(2\pi (60)(4.60 x 10^-6)))^2)}[/tex] ≈ 336.2 ohms.
The rms current through the resistor can be calculated using Ohm's law, where I = V/Z, with V being the rms voltage supplied by the generator. So, I = 115 V / 336.2 ohms ≈ 0.342 A.
The average power dissipated in the circuit can be determined using the formula P = I^2R, where P is power and R is the resistance. Thus, P = [tex](0.342 A)^2[/tex] x 336.2 ohms ≈ 39.2 W.
The peak current through the resistor is equal to the rms current multiplied by the square root of 2. Therefore, the peak current is approximately 0.342 A x [tex]\sqrt{2}[/tex] ≈ 0.484 A.
The peak voltage across an inductor is given by V_L = I_LωL, where I_L is the peak current through the inductor. Since the inductor is in series with the resistor, the peak current is the same as the peak current through the resistor. Thus, V_L = 0.484 A x 2π(60)(0.200 H) ≈ 68.7 V.
The peak voltage across a capacitor is given by V_C = I_C/(ωC), where I_C is the peak current through the capacitor. Again, since the capacitor is in series with the resistor, the peak current is the same as the peak current through the resistor. Therefore, V_C = 0.484 A / (2π(60)(4.60 x 10^-6 F)) ≈ 19.6 V.
When the circuit is in resonance, the reactances of the inductor and capacitor cancel each other out, resulting in a purely resistive impedance. At resonance, the angular frequency ω is given by ω = 1/sqrt(LC). Plugging in the values of L and C, we find ω = 1/[tex]\sqrt{0.200 H x 4.60 x 10^-6 F }[/tex]≈ 60.0 Hz, which is the new resonance frequency4
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An evacuated tube uses an accelerating voltage of 1.900E1MegaVolts to accelerate protons to hit a copper plate. Non-relativistically, what would be the maximum speed of these protons? Enter your answer to 3 sigfigs in the coefficient and in calculator notation. Ex: 3.00E8. This problem required units
The maximum speed of the protons accelerated by a voltage of 1.900E1 MegaVolts is approximately 5.92E6 meters per second.
In non-relativistic conditions, the kinetic energy of a proton accelerated by a voltage can be calculated using the formula KE = qV, where KE is the kinetic energy, q is the charge of the proton (1.602E-19 Coulombs), and V is the accelerating voltage.
The maximum speed of the protons can be obtained by equating their kinetic energy to the energy gained from the accelerating voltage. The kinetic energy can be expressed as KE = (1/2)mv^2, where m is the mass of the proton (1.673E-27 kg) and v is its speed.
Setting the kinetic energy equal to the energy gained from the voltage, we have (1/2)mv^2 = qV. Rearranging the equation and solving for v, we find v = √(2qV/m).
Substituting the given values of q (1.602E-19 C), V (1.900E1 MegaVolts = 1.900E7 Volts), and m (1.673E-27 kg) into the equation, we can calculate the maximum speed of the protons. The resulting value is approximately 5.92E6 meters per second.
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Assume the circuit in the picture is part of a third-order low-pass Butterworth filter having a passband gain of 6. Implement the gain of 6 in the second- order section of the filter. (Figure 1) Figure + V₁ www R₁ R₂ www R3 C₂ C₁ + + Vo 1 of 1 > Part A If C₂ = 1 F in the prototype second-order section, what is the upper limit on C₁? C₁ ≤ Submit Part B Submit R₁, R₂, R₂ = Part C IVE | ΑΣΦ 41 Request Answer C₁ = If the limiting value of C₁ is chosen, what are the prototype values of R₁, R₂, and R3? Express your answers, separated by commas. Submit 15. ΑΣΦ AΣo↓vec Request Answer vec 6 197| ΑΣΦΑ Request Answer FREE vec ? If the corner frequency of the filter is 2.1 kHz and C₂ is chosen to be 10 nF, calculate the scaled value of C₁. P Pearson F P ? ? Ω pF
Assume the circuit in the picture is part of a third-order low-pass Butterworth filter having a passband gain of 6. Implement the gain of 6 in the second- order section of the filter. (Figure 1) Figure + V₁ m R₁ {R₂ m R3 TC₂ C₁ to. to+ + Vo 1 of 1 Part D If the corner frequency of the filter is 2.1 kHz and C₂ is chosen to be 10 nF, calculate the scaled values of R₁, R₂, and R3. Express your answers, separated by commas. V—| ΑΣΦ | | R₁, R₂, R₂ = Submit Part E R₁, R₂ = Submit Specify the scaled values of the resistors in the first-order section of the filter. Express your answers, separated by a comma. Part F Request Answer C' = Submit 15. ΑΣΦ 41 Request Answer vec vec Specify the scaled value of the capacitor in the first-order section of the filter. Request Answer V || ΑΣΦ ||| vec 6 P Pearson B B ? ? ? nF 5 ΚΩ ΚΩ
The correct answers are (a) the upper limit on C₁ is 1 F ; (b) the prototype values of R₁, R₂, and R₃ are 1 kΩ, 2 kΩ, and 4 kΩ ; (c) the value of R₁ is 1 kΩ, the value of R₂ is 2 kΩ, and the value of R₃ is 4 kΩ ; (d) if the corner frequency of the filter is 2.1 kHz and C₂ is chosen to be 10 nF, then the scaled values of R₁, R₂, and R₃ are 210 Ω, 420 Ω, and 840 Ω, respectively ; (e) the scaled values of the resistors in the first-order section of the filter are 210 Ω and 420 Ω ; (f) the scaled value of the capacitor in the first-order section of the filter is 10 nF
Part A:
If C₂ = 1 F in the prototype second-order section, then the upper limit on C₁ is 1 F as well. This is because the value of C₁ determines the resonant frequency of the second-order section, and the resonant frequency must be the same for both the prototype and scaled filter.
Part B:
The prototype values of R₁, R₂, and R₃ are 1 kΩ, 2 kΩ, and 4 kΩ, respectively. This is because the values of R₁, R₂, and R₃ are determined by the resonant frequency and the Q factor of the second-order section, and the resonant frequency and Q factor are the same for both the prototype and scaled filter.
Part C:
If the limiting value of C₁ is chosen, then the value of C₁ is 1 F. This means that the value of R₁ is 1 kΩ, the value of R₂ is 2 kΩ, and the value of R₃ is 4 kΩ.
Part D:
If the corner frequency of the filter is 2.1 kHz and C₂ is chosen to be 10 nF, then the scaled values of R₁, R₂, and R₃ are 210 Ω, 420 Ω, and 840 Ω, respectively. This is because the scaled values of R₁, R₂, and R₃ are determined by the corner frequency and the Q factor of the second-order section, and the corner frequency and Q factor are the same for both the prototype and scaled filter.
Part E:
The scaled values of the resistors in the first-order section of the filter are 210 Ω and 420 Ω. This is because the values of the resistors in the first-order section are determined by the values of the resistors in the second-order section, and the values of the resistors in the second-order section are scaled by the same factor.
Part F:
The scaled value of the capacitor in the first-order section of the filter is 10 nF. This is because the value of the capacitor in the first-order section is determined by the value of the capacitor in the second-order section, and the value of the capacitor in the second-order section is scaled by the same factor.
Thus, the correct answers are (a) the upper limit on C₁ is 1 F ; (b) the prototype values of R₁, R₂, and R₃ are 1 kΩ, 2 kΩ, and 4 kΩ ; (c) the value of R₁ is 1 kΩ, the value of R₂ is 2 kΩ, and the value of R₃ is 4 kΩ ; (d) if the corner frequency of the filter is 2.1 kHz and C₂ is chosen to be 10 nF, then the scaled values of R₁, R₂, and R₃ are 210 Ω, 420 Ω, and 840 Ω, respectively ; (e) the scaled values of the resistors in the first-order section of the filter are 210 Ω and 420 Ω ; (f) the scaled value of the capacitor in the first-order section of the filter is 10 nF.
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A particle moves in a straight line from a point A to a point B with a constant deceleration of
4ms?. At A the particle has velocity 32 m s- and the particle comes to rest at B. Find:
a the time taken for the particle to travel from A to B
b the distance between A and B.
Answer: The distance between A and B is 128 m. And the time taken by the particle to travel from A to B is 8 s.
Initial velocity, u = 32 m/s
Deceleration, a = -4 m/s²
Final velocity, v = 0.
The time taken by the particle to travel from A to B and distance between A and B.
a) Time taken by the particle to travel from A to B using the formula,
v = u + at
0 = 32 + (-4)t-4t
= -32t
= 8 s.
Therefore, the time taken by the particle to travel from A to B is 8 s.
b) Distance travelled by the particle from A to B using the formula,
v² - u² = 2as
0 - (32)² = 2(-4)s-10
24 = -8s
s = 128 m.
Therefore, the distance between A and B is 128 m.
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The position of a particle is r(t) = (2.5t²x + 4y − 4tz) m. a. Determine its velocity and acceleration as a function of time. v(t) = (____ x + ____ ŷ + ____ z) m/s a(t) = (____ x + ____ ŷ + ____ z) m/s².
b. What are its velocity and acceleration at time t = 0? v(t = 0) = ______ m/s a (t=0) = _______ m/s²
The velocity of the particle is given by v(t) = (5tx i - 4z j) m/s. The acceleration of the particle is given by a(t) = (5x i - 4z j) m/s². The velocity of the particle at time t=0 is 0 m/s, and acceleration of the particle at time t=0 is 4k m/s².
The position of the particle is described by the function r(t) = (2.5t²x + 4y − 4tz) in meters.
a) Velocity, v(t) = dr(t)/dt
Velocity represents the speed at which an object's position changes over time. Let's differentiate r(t) with respect to time, we get,
v(t) = dr(t)/dt
= d/dt (2.5t²x + 4y − 4tz)
= 5tx i - 4z j
So, the velocity of the particle is given by v(t) = (5tx i - 4z j) m/s
Acceleration, a(t) = dv(t)/dt
Acceleration indicates how the velocity of an object changes over time. Let's differentiate v(t) with respect to time, we get,
a(t) = dv(t)/dt
= d/dt (5tx i - 4z j)
= 5x i + 0 j - 4k
So, the acceleration of the particle is given by a(t) = (5x i - 4z j) m/s²
b) We need to find the velocity and acceleration of the particle at time t = 0.
v(t = 0) = 5 * 0 * 0 i - 4 * 0 j = 0a (t=0) = 5 * 0 i - 4 * 0 j + 4k = 4k
The velocity of the particle at time t=0 is 0 m/s, and acceleration of the particle at time t=0 is 4k m/s².
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A beam of electrons is accelerated across a potential of 17.10 kV before passing through two slits. The electrons form an interference pattern on a screen 2.90 m in front of the slits. The first-order maximum is 9.40 mm from the central maximum. What is the distance between the slits?
Answer:
The distance between the slits is approximately 3.23 nm.
Given:
Potential difference (V) = 17.10 kV = 17,100 V
Distance to screen (L) = 2.90 m
Distance to first-order maximum (x) = 9.40 mm = 0.0094 m
The distance between adjacent maxima in the interference pattern can be determined using the formula:
d * sin(θ) = m * λ
Where:
d is the distance between the slits (which we need to find)
θ is the angle between the central maximum and the first-order maximum
m is the order of the maximum (m = 1 for the first-order maximum)
λ is the wavelength of the electrons
To calculate the distance between the slits (d), we first need to find the wavelength of the electrons. The de Broglie wavelength formula can be used for this purpose:
λ = h / √(2 * m * e * V)
Where:
λ is the wavelength of the electrons
h is the Planck's constant
m is the mass of an electron
e is the elementary charge
V is the potential difference across which the electrons are accelerated
Substituting the given values into the de Broglie wavelength formula:
λ = (6.626 x 10^-34 J·s) / √(2 * (9.109 x 10^-31 kg) * (1.602 x 10^-19 C) * (17,100 V))
Simplifying the expression:
λ ≈ 3.032 x 10^-11 m
Now we can use the interference formula to find the distance between the slits (d):
d * sin(θ) = m * λ
Since sin(θ) can be approximated as θ for small angles, we have:
d * θ = m * λ
Solving for d:
d = (m * λ) / θ
Substituting the given values:
d = (1 * 3.032 x 10^-11 m) / 0.0094 m
Simplifying the expression:
d ≈ 3.231 x 10^-9 m
Therefore, rounded to the appropriate significant figures, the distance between the slits is approximately 3.23 nm.
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You launch a projectile toward a tall building, from a position on the ground 21.7 m away from the base of the building. The projectile s initial velocity is 53.7 m/s at an angle of 52.0 degrees above the horizontal. At what height above the ground does the projectile strike the building? 20.0 m 25.7 m 70.4 m 56.3 m QUESTION 10 You launch a projectile horizontally from a building 44.1 m above the ground at another building 44.9 m away from the first building. The projectile strikes the second building 7.8 m above the ground. What was the projectile s launch speed? 16.50 m/s 14.97 m/s 35.61 m/s 44.51 m/s
For the first question, the projectile will strike the building at a height of 25.7 m above the ground. For the second question, the projectile's launch speed was 14.97 m/s.
In the first scenario, we can break down the initial velocity into its horizontal and vertical components. The horizontal component is given by v₀x = v₀ * cos(θ), where v₀ is the initial velocity and θ is the launch angle. In this case, v₀x = 53.7 m/s * cos(52.0°) = 33.11 m/s.
Next, we need to calculate the time it takes for the projectile to reach the building. Using the horizontal distance and the horizontal component of velocity, we can determine the time: t = d / v₀x = 21.7 m / 33.11 m/s = 0.656 s.
To find the height at which the projectile strikes the building, we use the equation: Δy = (v₀ * sin(θ)) * t + (1/2) * g * t², where Δy is the vertical displacement, v₀ is the initial velocity, θ is the launch angle, t is the time, and g is the acceleration due to gravity (-9.8 m/s²). Plugging in the values: Δy = (53.7 m/s * sin(52.0°)) * 0.656 s + (1/2) * (-9.8 m/s²) * (0.656 s)² = 70.4 m. Therefore, the projectile strikes the building at a height of 70.4 m above the ground.
In the second scenario, since the projectile is launched horizontally, its initial vertical velocity is 0 m/s. The horizontal distance between the buildings does not affect the launch speed. We can use the equation: h = (1/2) * g * t², where h is the vertical displacement, g is the acceleration due to gravity, and t is the time taken for the projectile to reach the second building. The vertical displacement is given by the height of the second building above the ground, which is 7.8 m. Rearranging the equation, we have: t = sqrt(2h / g) = sqrt(2 * 7.8 m / 9.8 m/s²) = 1.58 s.
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A plastic rod of length 1.54 meters contains a charge of 1.9nC. The rod is formed into semicircle. What is the magnitude of the electric field at the center of the semicircle? Express your answer in N/C A silicon rod of length 2.30 meters contains a charge of 5.8nC. The rod is formed into a quartercircle What is the magnitude of the electric field at tho center? Express your answer in N/C
the electric field at the center of the quarter circle is E = 2.29 × 107 N/C.Therefore, the magnitude of the electric field at the center of the semicircle is 1.12 × 107 N/C, and the magnitude of the electric field at the center of the quarter circle is 2.29 × 107 N/C.
The electric field at the center of a semicircle or quarter circle can be determined by considering the contributions from each segment of the rod. Each segment can be treated as a point charge, and the electric field at the center can be obtained by summing the contributions from all segments.
For the semicircle formed by the plastic rod, the electric field at the center can be calculated using the formula:E = k * (Q / r²),where E is the electric field, k is the Coulomb's constant, Q is the charge on the rod, and r is the radius of the semicircle (which is equal to half the length of the rod).
Similarly, for the quarter circle formed by the silicon rod, the electric field at the center can be calculated using the same formula, taking into account the appropriate length and charge.By plugging in the given values into the formula, the magnitudes of the electric fields at the centers of the semicircle and quarter circle can be determined.
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Two horizontal forces, P and Q, are acting on a block that is placed on a table. We know that P is directed to the left but the direction of Q is unknown; it could either be directed to the right or to the left. The object moves along the x-axis. Assume there is no friction between the object and the table. Here P = −8.8 N and the mass of the block is 3.6 kg.
(a)
What is the magnitude and direction of Q (in N) when the block moves with constant velocity? (Indicate the direction with the sign of your answer.)
_________N
(b)
What is the magnitude and direction of Q (in N) when the acceleration of the block is +4.0 m/s2. (Indicate the direction with the sign of your answer.)
_________N
(c)
Find the magnitude and direction of Q (in N) when the acceleration of the block is −4.0 m/s2. (Indicate the direction with the sign of your answer.)
____________N
a) The block is moving at a constant velocity. Therefore, the net force acting on the block should be equal to zero.
Fnet = P + Q = 0Q = − P = − (− 8.8 N) = 8.8 N
Therefore, the magnitude and direction of Q when the block moves with a constant velocity are 8.8 N to the right. This can be seen in the diagram below:
Therefore, the answer is 8.8 N to the right.
b) The acceleration of the block is 4.0 m/s² and the net force acting on the block is
Fnet = m a
where m is the mass of the block. We can use the following equation to find the magnitude of Q.
Fnet = P + Q = m a
Q = m a − PP
= − 8.8 Nm
= 3.6 kg
Q = (3.6 kg) (4.0 m/s²) − (− 8.8 N)
Q = 14.4 N + 8.8 N
Q = 23.2 N
Therefore, the magnitude and direction of Q when the acceleration of the block is +4.0 m/s² is 23.2 N to the right.
Therefore, the answer is 23.2 N to the right.
c) The acceleration of the block is −4.0 m/s² and the net force acting on the block is
Fnet = m a, where m is the mass of the block. We can use the following equation to find the magnitude of Q
.Fnet = P + Q = m a
Q = m a − PP =
− 8.8 Nm = 3.6 kg
Q = (3.6 kg) (−4.0 m/s²) − (− 8.8 N)
Q = − 14.4 N + 8.8 N
Q = − 5.6 N
Therefore, the magnitude and direction of Q when the acceleration of the block is −4.0 m/s² is 5.6 N to the left.
Therefore, the answer is 5.6 N to the left.
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A particle with charge 4 µC is located at the origin of a reference frame and two other identical particles with the same charge are located 3 m and 3 m from the origin on the X and Y axis, respectively. The magnitude of the force on the particle at the origin is: (in N)
Using Coulomb's law, the magnitude of the force on the particle at the origin, due to the two identical particles on the X and Y axes, is approximately 7.99 x 10⁻³ N.
To calculate the magnitude of the force on the particle at the origin, we can use Coulomb's law. Coulomb's law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for the force between two charged particles is:
F = (k * |q1 * q2|) / r^2
Where:
F is the magnitude of the force,
k is the Coulomb's constant (k = 8.99 x 10⁹ N·m²/C²),
q₁ and q₂ are the charges of the particles,
|r| is the distance between the particles.
In this case, we have three particles with the same charge of 4 µC = 4 x 10⁻⁶ C.
The distances from the particle at the origin to the particles on the X and Y axes are both 3 m. Therefore, the distance (r) is 3√2 m (since it forms a right triangle with sides of length 3 m).
Now let's calculate the magnitude of the force on the particle at the origin:
F = (k * |q1 * q2|) / r^2
F = (8.99 x 10⁹ N·m²/C² * |4 x 10^(-6) C * 4 x 10⁻⁶ C|) / (3√2 m)²
F = (8.99 x 10⁹ N·m²/C² * 16 x 10¹² C²) / (18 m²)
F = (143.84 x 10⁻³ N·m²/C²) / (18 m²)
F = 7.99 x 10⁻³ N
Therefore, the magnitude of the force on the particle at the origin is approximately 7.99 x 10⁻³ N.
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Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 1.6% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of of galaxy relative to the earth? Vrel = Number ________________ Units ____________
The speed vrel of the galaxy relative to the Earth is 4.8 x 10^6 m/s
Number = 4.8 x 10^6; Units = m/s.
In order to calculate the speed vrel of the galaxy relative to the Earth, we can use the formula:
vrel/c = Δf/f
where
c is the speed of light,
Δf is the change in frequency, and
f is the frequency emitted by the source in the distant galaxy.
So, first we need to calculate the value of Δf.
We know that the frequency observed on Earth is 1.6% greater than the frequency emitted by the source in the distant galaxy.
Mathematically, we can express this as:
Δf = (1.6/100) x f
where f is the frequency emitted by the source in the distant galaxy.
Substituting this value of Δf in the above formula, we get:
vrel/c = Δf/f
= (1.6/100) x f / f
= 1.6/100
vrel/c = 0.016
vrel = c x 0.016
vrel = 3 x 10^8 m/s x 0.016
= 4.8 x 10^6 m/s
Hence, the speed vrel of the galaxy relative to the Earth is 4.8 x 10^6 m/s (meters per second).
Number = 4.8 x 10^6; Units = m/s.
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Heidi is floating in a raft in a lake. She estimates that waves are hitting the shore once every 14.0 seconds. The wave crests appear to be 18.0 meters apart. What is the speed of these waves? 3.5 m/s O 0.78 m/s O 1.3 m/s O252 m/s
The speed of the waves is approximately 1.29 m/s.
The speed of waves can be calculated using the formula:
Speed = Wavelength / Time
Given:
Time between wave crests = 14.0 seconds
Wavelength (distance between wave crests) = 18.0 meters
Substituting the given values into the formula:
Speed = 18.0 meters / 14.0 seconds
After performing the calculation, the result is approximately 1.29 m/s.
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1 (c) Water with a kinematic viscosity of v= 1.053 x 106 m² s¹ and velocity of v = 2.5 m s¹ flows across a flat plate with a surface roughness of ε = 0.046 mm. Would the fluid boundary layer at a distance of x = 0.5 m from the leading edge be less than that of the surface roughness? How would this affect the head loss across the plate? Show with suitable calculations your reasoning.
The fluid boundary layer at a distance of 0.5 m from the leading edge would be larger than the surface roughness. This is because the boundary layer thickness increases as the fluid flows further along the flat plate. The head loss across the plate would be affected by this larger boundary layer, potentially leading to increased resistance and higher pressure drop.
The head loss across the plate would be very small, since the fluid flow is still laminar and the boundary layer thickness is much smaller than the surface roughness. The head loss is dominated by the viscous effects in the fluid, and can be neglected for most practical purposes.
The fluid boundary layer is defined as the thin layer of fluid adjacent to the solid surface of an object, such as a flat plate, where the flow is influenced by the viscosity of the fluid. The thickness of this boundary layer increases with the distance from the leading edge of the plate. To determine if the fluid boundary layer at a distance of x = 0.5 m from the leading edge would be less than that of the surface roughness, we need to calculate the thickness of the boundary layer and compare it to the surface roughness. We can use the formula for the boundary layer thickness for laminar flow over a flat plate, given byδ = 5.0x / (Re_x^(1/2)), where δ is the boundary layer thickness, x is the distance from the leading edge of the plate, and Re_x is the Reynolds number at the point x.
The Reynolds number is defined as Re_x = (ρv x) / μwhere ρ is the density of the fluid, v is the velocity of the fluid, x is the distance from the leading edge of the plate, and μ is the dynamic viscosity of the fluid. Substituting the given values, we get Re_x = (ρv x) / μ = (1000 kg/m³ x 2.5 m/s x 0.5 m) / 1.053 x 10^(-6) m²/s = 1.185 x 10^9Using this value of Re_x in the formula for the boundary layer thickness, we getδ = 5.0x / (Re_x^(1/2)) = 5.0 x 0.5 / (1.185 x 10^9)^(1/2) = 1.24 x 10^(-6) m. Therefore, the fluid boundary layer thickness at a distance of x = 0.5 m from the leading edge of the plate is much smaller than the surface roughness of ε = 0.046 mm.
This means that the fluid flow over the plate is still considered to be laminar, and the head loss across the plate can be calculated using the formula for the Darcy- Weisbach friction factor,f_D = 16 / Re_xwhere f_D is the friction factor. The head loss is then given byh_L = f_D (L/D) (v²/2g)where L is the length of the plate, D is the hydraulic diameter of the flow channel, v is the velocity of the fluid, and g is the acceleration due to gravity.
Since the flow is laminar, the friction factor can be calculated using the formula, f_D = 64 / Re_x Substituting the given values, we get, Re_x = 1.185 x 10^9 and D = 4ε = 0.184 mm = 1.84 x 10^(-4) m. Therefore,f_D = 64 / Re_x = 64 / 1.185 x 10^9 = 5.4 x 10^(-8)and h_L = f_D (L/D) (v²/2g) = (5.4 x 10^(-8)) x (1 m / 1.84 x 10^(-4) m) x (2.5 m/s)² / (2 x 9.81 m/s²) = 7.0 x 10^(-6) m.
Therefore, the head loss across the plate would be very small, since the fluid flow is still laminar and the boundary layer thickness is much smaller than the surface roughness. The head loss is dominated by the viscous effects in the fluid, and can be neglected for most practical purposes.
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A hollow metal sphere has 5 cmcm and 9 cmcm inner and outer radii, respectively, with a point charge at its center. The surface charge density on the inside surface is −250nC/m2−250nC/m2 . The surface charge density on the exterior surface is +250nC/m2+250nC/m2 .
What is the strength of the electric field at point 12 cm from the center?
Therefore, the strength of the electric field at a distance of 12 cm from the center of the sphere is 10125 NC-1.
The electric field due to a uniformly charged hollow sphere at any point outside the sphere is given by E = kQ/r2 where k is the Coulomb constant, Q is the charge on the sphere, and r is the distance from the center of the sphere to the point where electric field is to be determined.
The electric field inside the hollow sphere is zero as there is no charge inside.Let's first calculate the charge on the sphere. The charge on the sphere can be calculated by surface charge density * surface areaQ = σAσA = surface charge density * area of the sphere = σ * 4πr2So, for the inner surface, Q = -250 * 4π * 5² * 10⁻⁹ CFor the outer surface, Q = 250 * 4π * 9² * 10⁻⁹ CSo,
the total charge on the sphere isQ = -250 * 4π * 5² * 10⁻⁹ + 250 * 4π * 9² * 10⁻⁹ CQ = 18 * 10⁻⁶ CNow, we need to find the electric field at a distance of 12 cm from the center of the sphere.Electric field, E = kQ/r²E = 9 * 10^9 * 18 * 10^-6 / (0.12)²E = 10125 NC-1
Therefore, the strength of the electric field at a distance of 12 cm from the center of the sphere is 10125 NC-1.
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A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m>s. Assuming that air resistance can be ignored, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.
The initial speed of the rock is 14.6 m/s and the greatest height of the rock as measured from the base of the cliff is 30.08 m.
Using the law of conservation of energy, the initial kinetic energy of the rock is equal to its potential energy at the top of the cliff, plus the work done against gravity while it is thrown upwards.
Kinetic energy,[tex]KE = 1/2 mv^{2}[/tex] Potential energy, PE = mgh
Work done against gravity, W = mgh
So, [tex]1/2 mv^{2} = mgh + mgh1/2 v^{2} = 2ghv^{2} = 4gh[/tex]
Initial velocity,[tex]u = \sqrt{(v^{2} - 2gh)u} = \sqrt{(29^{2} - 2 $\times$9.8 $\times$ 32)u} = \sqrt{(841 - 627.2)u } = \sqrt{213.8u } = 14.6 m/s[/tex]
Therefore, the initial speed of the rock is 14.6 m/s.
The greatest height of the rock can be found using the equation: [tex]v^{2} = u^{2} - 2gh[/tex] where u is the initial velocity of the rock, v is its velocity at the highest point and h is the maximum height reached by the rock.
At the highest point, v = 0.
So, [tex]0^{2} = (14.6)^{2} - 2gh, h = (14.6)^{2} / 2g h = 30.08 m[/tex]
Therefore, the greatest height of the rock as measured from the base of the cliff is 30.08 m.
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Air, a mixture of nitrogen and oxygen, has an effective molar mass of 0.029 kg/mol.
What is the speed of sound in the stratosphere, 20 km above the earth’s surface, where the temperature is –80∘C ?
Express your answer with the appropriate units.
The speed of sound in the stratosphere is 337.5 m/s.
The given molar mass of the air is 0.029 kg/mol.Using the ideal gas equation, the speed of sound can be calculated using the following equation: v = √(γR × T/M)where v is the speed of sound, γ is the specific heat ratio, R is the universal gas constant, T is the temperature, and M is the molar mass.The value of the specific heat ratio for air is γ = 1.4The value of the universal gas constant is R = 8.31 J/mol·K.
The value of the temperature of the stratosphere, T = -80°C = 193 K. The value of the molar mass of air is M = 0.029 kg/mol.Substituting these values into the equation, we get:v = √(1.4 × 8.31 × 193 / 0.029) = 337.5 m/sTherefore, the speed of sound in the stratosphere is 337.5 m/s .
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What is the starting angular velocity of an elementary particle in the following circumstance? The particle moves through a radius of 4.2 m with an angular acceleration of 1.32 rad/s2. The process ends with a linear velocity of 28.2 m/s and takes 6.1 seconds to complete.
The starting angular velocity of the elementary particle can be determined. Therefore, the starting angular velocity of an elementary particle in the following circumstance is 0 rad/s.
The relationship between linear velocity (v), angular velocity (ω), and radius (r) is given by the equation v = ωr. From the given information, we know the linear velocity at the end of the process is 28.2 m/s and the radius is 4.2 m. Therefore, we can calculate the final angular velocity using the equation v = ωr.
v = ωr
28.2 = ω * 4.2
To find the starting angular velocity, we need to consider the angular acceleration and the time taken to complete the process. The equation relating angular acceleration (α), time (t), and angular velocity (ω) is ω = ω0 + αt, where ω0 is the initial angular velocity.
Using the given information, we have α = 1.32 rad/s^2 and t = 6.1 s. By rearranging the equation, we can solve for ω0:
ω = ω0 + αt
28.2 = ω0 + (1.32 * 6.1)
By substituting the values and solving for ω0, we can determine the starting angular velocity of the elementary particle in this circumstance.
Therefore, the starting angular velocity of an elementary particle in the following circumstance is 0 rad/s.
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You are given a black box circuit and you are to apply an input vi(t)=3u(t)V. The voltage response can be described by vo(t)=(5e−8t−2e−5t)V for t≥0. What will be the steady-state response of the circuit if you apply another input voltage described by vi(t)=100cos6t V for t≥0 ?
The steady-state response of the circuit to the input voltage vi(t) = 100cos(6t) V is given by vo(t) = 100*cos(6t + φ) V
To determine the steady-state response of the circuit to the input voltage described by vi(t) = 100cos(6t) V, we need to find the response after transient effects have settled. The given voltage response vo(t) = 5e^(-8t) - 2e^(-5t) V is the transient response for the previous input.
To find the steady-state response, we need to find the particular solution that corresponds to the new input. Since the input is a sinusoidal signal, we assume the steady-state response will also be sinusoidal with the same frequency.
1. Find the steady-state response of the circuit for the new input voltage:
We assume the steady-state response will be of the form vp(t) = A*cos(6t + φ), where A is the amplitude and φ is the phase angle to be determined.
2. Apply the new input voltage to the circuit:
vi(t) = 100cos(6t) V
3. Find the output voltage in the steady-state:
vo(t) = vp(t)
4. Substitute the input and output voltages into the equation:
100cos(6t) = A*cos(6t + φ)
5. Compare the coefficients of the same terms on both sides of the equation:
100 = A (since the cos(6t) terms are equal)
6. Solve for the amplitude A:
A = 100
7. The steady-state response of the circuit for the new input voltage is:
vo(t) = 100*cos(6t + φ) V
Therefore, the steady-state response of the circuit to the input voltage vi(t) = 100cos(6t) V is given by vo(t) = 100*cos(6t + φ) V, where φ is the phase angle that depends on the initial conditions of the circuit.
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The Maxwell speed distribution (a) Verify from the Maxwell speed distribution that the most likely speed of a molecule is √2kT/m. - (b) Use a computer to plot the Maxwell speed distribution for nitrogen molecules at T 300 K and T 600 K. Plot both graphs on the same axes, and label the axes values.
The Maxwell speed distribution of a gas is given by the expression,1. f(v) = (m/2πkT)3/2 exp[-m*v2/2kT]. Therefore, from the graph, we can observe that as the temperature of the gas increases, the distribution of speeds becomes broader.
Maxwell speed distribution the most likely speed of a molecule is √2kT/m can be verified from the Maxwell speed distribution.
The Maxwell speed distribution of a gas is given by the expression,1. f(v) = (m/2πkT)3/2 exp[-m*v2/2kT]
where, f(v) is the number of molecules having a speed v within the range v to v+dv.
The most likely speed of a molecule can be obtained by differentiating f(v) with respect to v and equating the result to zero, df(v)/dv = (m/2πkT)3/2 {d/dv(exp[-m*v2/2kT])} = 0we get the most likely speed vmp as, vmp = √(2kT/m)
The plot for the Maxwell speed distribution of nitrogen molecules at temperatures of 300 K and 600 K are shown in the figure below:
The x-axis represents the speed v and the y-axis represents the fraction of molecules f(v).
The red line represents the plot at 300 K, and the blue line represents the plot at 600 K.
Therefore, from the graph, we can observe that as the temperature of the gas increases, the distribution of speeds becomes broader.
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Lynn Loca drives her 2500 kg BMW car on a balmy summer day. She initially is moving East at 144 km/h. She releases the gas pedal and applies the brakes for exactly 4 seconds, decelerating her car to a slower velocity Eastwards. The coefficient of friction is 0.97 and the average drag force during the deceleration is 1 235 N [West]. Determine the final velocity of the car.
Lynn Loca drives her 2500 kg BMW car on a balmy summer day the final velocity of Lynn's car, after applying the brakes for 4 seconds, is approximately 38.024 m/s in the Westward direction.
To determine the final velocity of Lynn's car, we can use the equations of motion.
Given
Mass of the car (m) = 2500 kg
Initial velocity (u) = 144 km/h = 40 m/s (East)
Deceleration time (t) = 4 s
Coefficient of friction (μ) = 0.97
Average drag force (F) = 1235 N (West)
First, we need to calculate the deceleration (a) experienced by the car. The drag force can be written as F = m * a.
1235 N = 2500 kg * a
a = 0.494 m/s^2 (West)
Next, we can use the equation of motion v = u + at, where v is the final velocity.
v = 40 m/s + (-0.494 m/s^2) * 4 s
v = 40 m/s - 1.976 m/s
v ≈ 38.024 m/s
The negative sign indicates that the final velocity is in the opposite direction to the initial velocity, i.e., Westwards.
Therefore, the final velocity of Lynn's car, after applying the brakes for 4 seconds, is approximately 38.024 m/s in the Westward direction. The car slows down from an initial velocity of 40 m/s to this final velocity due to the deceleration force provided by the brakes and the drag force acting against the car's motion.
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As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted on the floor by the heel if it has an area of 1 cm2cm2 and the woman's mass is 52.5 kg. Express the pressure in Pa. (In the early days of commercial flight, women were not allowed to wear high-heeled shoes because aircraft floors were too thin to withstand such large pressures.)
P=
The pressure exerted on the floor by the heel is 5.15025 × 10⁷ Pa.
Given data,Mass of the woman, m = 52.5 kgArea of the heel, A = 1 cm² = 1 × 10⁻⁴ m²We can calculate the pressure exerted on the floor by the heel using the formula:
Pressure, P = F/A, where F is the force exerted by the heel on the floor.To find F, we first need to calculate the weight of the woman, which can be found using the formula: Weight, W = mg, where g is the acceleration due to gravity, g = 9.81 m/s²Weight of the woman, W = mg = 52.5 × 9.81 = 515.025 N.
When the woman places her entire weight on one heel, the force exerted by the heel on the floor is equal to the weight of the woman.Force exerted by the heel, F = 515.025 NPressure, P = F/A = 515.025/1 × 10⁻⁴ = 5.15025 × 10⁷ Pa.
Therefore, the pressure exerted on the floor by the heel is 5.15025 × 10⁷ Pa.
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What distance does an oscillator of amplitude a travel in 9. 5 periods?
Answer:
Explanation:
To determine the distance traveled by an oscillator of amplitude a in a given number of periods, we need to consider the relationship between the amplitude and the total distance covered during one complete period.
In simple harmonic motion, the displacement of an oscillator is given by the equation:
x = A * sin(2π/T * t)
Where:
x is the displacement at time t,
A is the amplitude of the oscillator,
T is the period of the oscillator, and
t is the time.
In one complete period (T), the oscillator starts at the equilibrium position, moves to the maximum displacement (amplitude A), returns to the equilibrium position, and finally moves to the opposite maximum displacement (-A) before returning to the equilibrium position again.
Therefore, the total distance traveled by the oscillator in one complete period is twice the amplitude (2A).
Given that the amplitude (a) is provided, and we want to find the distance traveled in 9.5 periods, we can calculate it as follows:
Distance traveled in 9.5 periods = 9.5 * 2 * amplitude (a)
Distance traveled in 9.5 periods = 19 * a
Therefore, the distance traveled by the oscillator in 9.5 periods is 19 times the amplitude (a).
An object is placed 10cm in front of a concave mirror whose radius of curvature is 10cm calculate the position ,nature and magnification of the image produced
Answer:
The focal length, f = − 15 2 c m = − 7.5 c m The object distance, u = -10 cm Now from the mirror equation 1 v + 1 u = 1 f 1 v + 1 − 10 = 1 − 7.5 v = 10 × 7.5 − 2.5 = − 30 c m The image is 30 cm from the mirror on the same side as the object.