The ratio of increase in their lengths is 2:1. Answer: 2:1.
Let the length and radius of the first wire be 2L and 2r and the length and radius of the second wire be L and r.According to the question, both wires are made up of the same metal and equal loads are applied to both wires.We can use Young's Modulus to calculate the ratio of the increase in their lengths. Young's modulus, also known as the modulus of elasticity, is a material property that relates the stress (force per unit area) to the strain (change in length per unit length) in a material.
Mathematically, it is given as:E = stress/strainE = FL/ArWhere,F = load appliedL = original length of the wireA = cross-sectional area of the wirer = radius of the wireLet the increase in length of both wires be ΔL and Δl for the first and second wire, respectively. Then,ΔL = FL/ArEAndΔl = Fl/arEThe ratio of increase in their lengths is:ΔL/Δl= (FL/Ar) / (Fl/arE)= 2L / L= 2/1Therefore, the ratio of increase in their lengths is 2:1. Answer: 2:1
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A.5.0 mH inductor is connected in parallel with a variable capacitor. The capacitor can be varied from 140pF to 380pF What is the minimum oscillation frequency for this circuit? Express your answer with the appropriate units.Part B What is the maximum oscillation frequency for this circuit? Express your answer with the appropriate units.
A.5.0 mH inductor is connected in parallel with a variable capacitor. the minimum oscillation frequency for this circuit is approximately 1.06 MHz. the minimum oscillation frequency for this circuit is approximately 1.06 MHz.
To determine the minimum and maximum oscillation frequencies for the circuit consisting of a 5.0 mH inductor and a variable capacitor ranging from 140 pF to 380 pF, we can use the formula for the resonant frequency of an LC circuit:
f = 1 / (2π√(LC))
The resonant frequency, f, is the frequency at which the circuit exhibits maximum oscillation or resonance. The minimum oscillation frequency occurs when the capacitance is at its maximum value, and the maximum oscillation frequency occurs when the capacitance is at its minimum value.
For the minimum oscillation frequency:
C = 380 pF = 380 × 10^(-12) F
L = 5.0 mH = 5.0 × 10^(-3) H
Substituting these values into the formula, we get:
f_min = 1 / (2π√(5.0 × 10^(-3) H × 380 × 10^(-12) F))
= 1 / (2π√(1.9 × 10^(-15) H·F))
≈ 1.06 MHz
Therefore, the minimum oscillation frequency for this circuit is approximately 1.06 MHz.
For the maximum oscillation frequency, we use the minimum value of the capacitor:
C = 140 pF = 140 × 10^(-12) F
Substituting this value into the formula, we get:
f_max = 1 / (2π√(5.0 × 10^(-3) H × 140 × 10^(-12) F))
= 1 / (2π√(7.0 × 10^(-16) H·F))
≈ 2.04 MHz
Therefore, the minimum oscillation frequency for this circuit is approximately 1.06 MHz.
In summary, the minimum oscillation frequency is approximately 1.06 MHz, occurring when the capacitor is at its maximum value of 380 pF. The maximum oscillation frequency is approximately 2.04 MHz, occurring when the capacitor is at its minimum value of 140 pF. These frequencies represent the resonant frequencies at which the LC circuit will exhibit maximum oscillation or resonance.
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Two waves on one string are described by the wave functions
y1= 2.05 cos(3.05x − 1.52t)
y2= 4.54 sin(3.31x − 2.39t)
where x and y are in centimeters and t is in seconds. (Remember that the arguments of the trigonometric functions are in radians.)
(a) Find the superposition of the waves y_1 + y_2y1+y2 at x = 1.0, t = 0.0 s.
Two waves on one string are described by the wave functions y1= 2.05 cos(3.05x − 1.52t),y2= 4.54 sin(3.31x − 2.39t)where x and y are in centimeters and t is in seconds.The superposition of the waves y1 + y2 at x = 1.0 cm and t = 0.0 s is approximately 2.099968 cm.
To find the superposition of the waves at a specific point (x, t), we need to add the values of the two wave functions at that point.
Given:
y1 = 2.05 cos(3.05x - 1.52t)
y2 = 4.54 sin(3.31x - 2.39t)
x = 1.0 cm
t = 0.0 s
We can substitute the given values into the wave functions and perform the addition.
y1 + y2 = 2.05 cos(3.05x - 1.52t) + 4.54 sin(3.31x - 2.39t)
Substituting x = 1.0 cm and t = 0.0 s:
y1 + y2 = 2.05 cos(3.05(1.0) - 1.52(0.0)) + 4.54 sin(3.31(1.0) - 2.39(0.0))
y1 + y2 = 2.05 cos(3.05) + 4.54 sin(3.31)
Using a calculator, evaluate the cosine and sine functions:
y1 + y2 ≈ 2.05 * 0.999702 + 4.54 * 0.011432
y1 + y2 ≈ 2.048031 + 0.051937
y1 + y2 ≈ 2.099968
Therefore, the superposition of the waves y1 + y2 at x = 1.0 cm and t = 0.0 s is approximately 2.099968 cm.
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A circular loop of wire has a radius of 0.025 m and a resistance of 3.0Ω. It is placed in a 1.6 T magnetic field which is directed in through the loop as shown. a) Calculate the change in magnetic flux in the circular loop when the magnetic field turned off. [3 marks] b) If the circular loop has 140 turns, what is the emf induced in the loop at t=0.18 s? [2 marks] c) What is the current induced in the loop? [2 marks] d) State the direction of the current induced in the loop
a) The change in magnetic flux when the magnetic field is turned off is 0.08 Wb. b) The induced emf in the loop at t=0.18 s is 0.672 V. c) The induced current in the loop is 0.224 A. d) The current induced in the loop flows counterclockwise.
a) Change in magnetic flux is given by:ΔΦ = Φ₂ - Φ₁Φ₂ is the final magnetic flux, Φ₁ is the initial magnetic flux. Given that the magnetic field is turned off, the final magnetic flux Φ₂ becomes zero. We can calculate the initial magnetic flux Φ₁ using the formula:Φ₁ = BA. Where B is the magnetic field strength, and A is the area of the circular loop. Substituting the given values, we get:Φ₁ = πr²B = π(0.025)² (1.6)Φ₁ = 1.25 x 10⁻³ Wb. Therefore, the change in magnetic flux is:ΔΦ = Φ₂ - Φ₁ΔΦ = 0 - 1.25 x 10⁻³ΔΦ = -1.25 x 10⁻³ Wb)
The emf induced in the circular loop is given by the formula:ε = -N (ΔΦ/Δt). Substituting the given values, we get:ε = -140 (-1.25 x 10⁻³/0.18)ε = 10.97 Vc) The current induced in the circular loop is given by the formula: I = ε/R. Substituting the given values, we get: I = 10.97/3.0I = 3.66 Ad) The direction of the current induced in the circular loop can be determined by Lenz's law, which states that the induced current will flow in a direction such that it opposes the change in magnetic flux that produced it. In this case, when the magnetic field is turned off, the induced current will create a magnetic field in the opposite direction to the original field, to try to maintain the flux. Therefore, the current will flow in a direction such that its magnetic field points into the loop.
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A sharp image is located 321 mm behind a 214 mm focal-length converging lens. Find the object distance. Give answer in mm. Unanswered ⋅3 attempts left How far apart are an object and an image formed by a 97 cm lens, if image is 2.6 larger than the object and real? Give answer in cm. Unanswered ⋅3 attempts left How far apart are an object and an image formed by a 97 cm lens, if image is 2.6 larger than the object and virtual? Give answer in cm. Unanswered ⋅3 attempts left The near and far point of some person are 10.9 cm and 22.0 respectively. She got herself the perfect contacts for driving. What is the near point of this person with lens in place? Give answer is cm.
Q1) A sharp image is located 321 mm behind a 214 mm focal-length converging lens.
Find the object distance.
Give answer in mm.
Given, f = 214 mmv = -321 mm
Using the lens formula,1/f = 1/v - 1/u
Where, u is the object distance.
Substituting the given values, we get
1/214 = 1/-321 - 1/u
Multiplying both sides by -214*-321*u, we get-u = 214 * -321 / (214 - -321)u = -4596 mm
The object distance is -4596 mm.
Q2) How far apart are an object and an image formed by a 97 cm lens, if the image is 2.6 larger than the object and real? Give the answer in cm.
Given, f = 97 cm
Image is real and 2.6 times larger than the object.
u = ?
Using magnification formula, magnification, m = -v/u where, magnification m = 2.6for real images, v is negative and for virtual images, v is positive.
Substituting the given values,2.6 = -v/u
Since the object and image distance are far apart, v = u + d Where d is the separation between the object and image substituting v in terms of u,2.6 = -(u + d)/u Simplifying the above expression, we get u = -36.154 cm
Therefore, the object and image distance is 36.154 cm apart.
Q3) How far apart are an object and an image formed by a 97 cm lens, if the image is 2.6 larger than the object and virtual? Give the answer in cm.
Given,
f = 97 cm Image is virtual and 2.6 times larger than the object.
u = ?
Using magnification formula, magnification, m = v/where, magnification m = 2.6for real images, v is negative and for virtual images, v is positive. Substituting the given values,2.6 = v/u Since the object and image distance are far apart, v = -(u + d)Where d is the separation between the object and image
Substituting v in terms of u,2.6 = (u + d)/u
Simplifying the above expression, we get u = 30.4 cm
Therefore, the object and image distance is 30.4 cm apart.
Q4) The near and far point of some person are 10.9 cm and 22.0, respectively. She got herself the perfect contacts for driving. What is the near point of this person with the lens in place? Give the answer is cm.
Given,v1 = 10.9 cmv2 = 22.0 cm
Using the formula, lens formula,1/f = 1/v1 - 1/u
Where, u is the distance of the lens from the near point of the eye.
Substituting the given values, we get1/f = 1/10.9 - 1/u
Simplifying the above expression, we get u = -35.5 cm
Using the formula, lens formula,1/f = 1/v2 - 1/u Where, u is the distance of the lens from the far point of the eye.
Substituting the given values, we get1/f = 1/22 - 1/u
Simplifying the above expression, we get u = 77 cm
The near point of the person with the lens in place is at a distance of
35.5 cm.
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Given a three-phase AC power network with a load, which consumes 100 MW with a power factor of 0.8 (lagging). Three capacitors with equal values are connected in star formation across the load to improve the power factor to 0.96 (leading). Calculate the reactive power supplied by the three capacitors
Active power consumed by the load P = 100 MW P.F of the load cos φ = 0.8 (lagging)
P.F of the load after connecting capacitors cos φ2 = 0.96 (leading)
The formula to calculate the reactive power is
Q = P(tan φ1 - tan φ2) Where, Q = Reactive power required by capacitors P = Active power consumed by the load
cos φ1 = Power factor of the load before adding capacitors
cos φ2 = Power factor of the load after adding capacitors
tan φ1 = √(1 - cos²φ1)/cos φ1
tan φ1 = √(1 - 0.8²)/0.8 = 0.6
tan φ2 = √(1 - cos²φ2)/cos φ2
tan φ2 = √(1 - 0.96²)/0.96 = 0.4
Therefore, Q = 100 × (0.6 - 0.4) = 20 MW
Thus, the reactive power supplied by the three capacitors is 20 MW.
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Which is more efficient, a toaster that converts 95% of the
energy it receives to heat or an incandescent light bulb which ALSO
converts 95% of its energy to heat? Explain
Both the toaster and the incandescent light bulb have the same energy conversion efficiency of 95% in terms of heat. However, the toaster is more efficient in terms of utility because it directly provides heat for toasting, while the light bulb primarily produces light and converts a smaller portion of energy into heat.
Both the toaster and the incandescent light bulb convert 95% of the energy they receive into heat. However, the key difference lies in their intended purpose and utility.
A toaster is specifically designed to generate heat for toasting bread or other food items. Its primary function is to convert electrical energy into heat energy efficiently.
Therefore, the 95% energy conversion efficiency of the toaster is directly utilized for its intended purpose, making it highly efficient in terms of utility.
On the other hand, an incandescent light bulb is primarily designed to produce light, with heat being a byproduct of its operation. While it is true that 95% of the energy consumed by the incandescent light bulb is converted into heat, the primary function of the light bulb is to emit visible light.
The heat generated by the bulb is often considered a waste product in this context, as it does not serve a direct purpose for illumination. In conclusion, while both the toaster and the incandescent light bulb have the same energy conversion efficiency of 95% in terms of heat.
The toaster is more efficient in terms of utility because it directly provides the desired heat for toasting, whereas the incandescent light bulb primarily produces light and the heat generated is considered a byproduct.
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What does it mean when white light is diffracted and at a particular location the color seen is blue?
Answer:
When white light is diffracted and at a particular location , the color seen is blue, this means that blue color is reflected while all other colors are absorbed or diffracted . This phenomenon is due to the absorbance of wavelengths of all other colors except blue.
Explanation:
The G Key below middle C on a piano keyboard vibrates with a
frequency of 390Hz. Determine the period of vibration.
The period of vibration of the G key below middle C on a piano keyboard is 0.0026 seconds.
Given that the frequency of the G key below middle C on a piano keyboard vibrates is 390 Hz.To determine the period of vibration, the formula to use is given as;T = 1/fWhere;T = period of vibrationf = frequency of the vibrationUsing the formula,T = 1/390Period of vibration T = 0.0026 secondsWe can also say that the G key below middle C on a piano keyboard vibrates with a period of 0.0026 seconds.Therefore, the period of vibration of the G key below middle C on a piano keyboard is 0.0026 seconds.
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The smaller the resistance in an LRC circuit, the greater the resonance peak current. True False
False. The smaller the resistance in an LRC (inductor-resistor-capacitor) circuit, the lower the resonance peak current.
In an LRC circuit, resonance occurs when the angular frequency of the driving AC source matches the natural frequency of the circuit. At resonance, the current in the circuit is maximized. The resonance frequency can be calculated using the formula [tex]\omega = \frac{1}{\sqrt{LC}}[/tex], where L is the inductance and C is the capacitance in the circuit.
However, the resistance in the circuit affects the behavior of the current at resonance. The presence of resistance causes energy dissipation and leads to a decrease in the resonance peak current. This is due to the fact that the resistance limits the flow of current and dissipates some of the energy.
As the resistance decreases in the LRC circuit, the energy dissipation decreases, resulting in a smaller loss of energy. Consequently, the resonance peak current increases as the resistance decreases. Therefore, the statement that the smaller the resistance in an LRC circuit, the greater the resonance peak current is false.
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This time we have a non-rotating space station in the shape of a long thin uniform rod of mass 4.72 x 10^6 kg and length 1491 meters. Small probes of mass 9781 kg are periodically launched in pairs from two points on the rod-shaped part of the station as shown, launching at a speed of 2688 m/s with respect to the launch points, which are each located 493 m from the center of the rod. After 11 pairs of probes have launched, how fast will the station be spinning?
3.73 rpm
1.09 rpm
3.11 rpm
1.56 rpm
The correct option is c. After launching 11 pairs of probes from the non-rotating space station, the station will be at a spinning rate of approximately 3.11 rpm (revolutions per minute).
To determine the final spin rate of the space station, we can apply the principle of conservation of angular momentum. Initially, the space station is not spinning, so its initial angular momentum is zero. As the pairs of probes are launched, they carry angular momentum with them due to their mass, velocity, and distance from the center of the rod.
The angular momentum carried by each pair of probes can be calculated as the product of their individual masses, velocities, and distances from the center of the rod. The total angular momentum contributed by the 11 pairs of probes can then be summed up.
Using the principle of conservation of angular momentum, the total angular momentum of the space station after the probes are launched should be equal to the sum of the angular momenta carried by the probes. From this, we can determine the final angular velocity of the space station.
Converting the angular velocity to rpm (revolutions per minute), we find that the space station will be spinning at a rate of approximately 3.11 rpm after launching 11 pairs of probes.
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An electron with a speed of 5x10 m/s experiences an acceleration of
magnitude 2x10" m/s° in a magnetic field of strength 2.6 T. What is
the angle between the velocity and magnetic field?
2. An electron is shot with a horizontal initial velocity in an upward
uniform magnetic field of 1.5 mT. It moves in a circle in the field.
a. (a) Does it move clockwise or counterclockwise?
b. (b) How long does each orbit take?
c. (c) If the radius of the circle is 1.3 cm then what is the speed of
the electron?
3. A long, straight wire on the x axis carries a current of 3.12 A in the
positive x direction. The magnetic field produced by the wire
combines with a uniform magnetic field of 1.45x10°that points in the
positive z direction. (a) Is the net magnetic field of this system equal
to zero at a point on the positive y axis or at a point on the negative y
axis? Explain. (b) Find the distance from the wire to the point where
the field vanishes.
4. A solenoid has a circular cross-section with a 3 cm radius, a length of
80 cm and 300 turns. It carries a current of 5 A. What is the magnetic
field strength inside the solenoid?
An electron with a speed of 5x10 m/s experiences an acceleration of magnitude 2x10" m/s° in a magnetic field of strength 2.6 T. the angle using a calculator, we find the angle to be approximately 0.001 radians. the speed of the electron to be approximately 2.42x10^6 m/s.
1. To find the angle between the velocity and magnetic field for an electron, we can use the formula:
a = (qvB) / m,
where a is the acceleration, q is the charge of the electron, v is the velocity, B is the magnetic field strength, and m is the mass of the eletron.
Given:
v = 5x10^6 m/s,
a = 2x10^6 m/s^2,
B = 2.6 T.
The charge of an electron is q = -1.6x10^-19 C, and the mass of an electron is m = 9.11x10^-31 kg.
Substituting the values into the formula:
2x10^6 = (1.6x10^-19)(5x10^6)(2.6) / (9.11x10^-31).
Simplifying the equation, we can solve for the magnitude of the angle:
angle = arctan(2x10^6 * 9.11x10^-31 / (1.6x10^-19 * 5x10^6 * 2.6)).
Calculating the angle using a calculator, we find the angle to be approximately 0.001 radians.
2. (a) Since the electron is moving in a circle in the magnetic field, its motion is perpendicular to the magnetic field. According to the right-hand rule, the direction of the force experienced by a negative charge moving perpendicular to a magnetic field is opposite to the direction of the field. Therefore, the electron moves counterclockwise.
(b) The time taken for each orbit can be calculated using the formula:
T = (2πm) / |q|B),
where T is the time period, m is the mass of the electron, q is the charge of the electron, and B is the magnetic field strength.
Given:
m = 9.11x10^-31 kg,
q = -1.6x10^-19 C,
B = 1.5 mT = 1.5x10^-3 T.
Substituting the values into the formula:
T = (2π * 9.11x10^-31) / (|-1.6x10^-19| * 1.5x10^-3).
Calculating the time period using a calculator, we find T to be approximately 3.77x10^-8 seconds.
(c) The speed of the electron can be determined using the formula for the centripetal force:
F = (mv^2) / r,
where F is the magnetic force acting on the electron, m is the mass of the electron, v is the velocity of the electron, and r is the radius of the circle.
Given:
m = 9.11x10^-31 kg,
v = unknown,
r = 1.3 cm = 1.3x10^-2 m.
The magnetic force acting on the electron is given by the equation:
F = |q|vB,
where q is the charge of the electron and B is the magnetic field strength.
Substituting the values and equations into the formula:
|q|vB = (mv^2) / r.
Simplifying the equation, we can solve for the speed of the electron:
v = (rB) / |q|.
Substituting the values:
v = (1.3x10^-2)(1.5x10^-3) / |-1.6x10^-19|.
Calculating the speed using a calculator, we find the speed of the electron to be approximately 2.42x10^6 m/s.
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Electric room heaters use a concave mirror to reflect infrared (IR) iradiation from hot coils. Note that IR follows the siume law of reflection as visible light.
Find the magnification of the heater element, given that the mirror has a radius of curvature of 48 cm and produces an image of the coils 3.2 m away from the mirror.
M = ______________
the magnification of the heater element is 0.5.
radius of curvature (r) of the mirror = 48 cm
Image distance (v) = 3.2 m
Focal length (f) = r/2 = 48/2 = 24 cm
According to mirror formula:1/v + 1/u = 1/f
Where,
u is object distance.
In this case, u = -f [since the object is placed at the focus]
1/v = 1/f - 1/u=> 1/v = 1/24 + 1/24=> 1/v = 1/12=> v = 12 m
Magnification (M) is given as:
Magnification M = -v/u=> M = -12/-24= 0.5
So, the magnification of the heater element is 0.5.
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Using the Bohr model, calculate the speed of an electron in the ground state, and express the speed in terms of c. During our derivation of the allowed atomic energies in the Bohr model, we used nonrelativistic formulas-was this assumption justifiable?
The speed of an electron in the ground state of hydrogen, according to the Bohr model, is c/137.
The assumption of using nonrelativistic formulas in the Bohr model is justifiable for low-energy and low-velocity systems, but it becomes less accurate and applicable as the electron's speed approaches the speed of light.
In the Bohr model, the speed of an electron in the ground state can be calculated using the formula:
v = αc / n
where v is the speed of the electron, α is the fine structure constant (approximately 1/137), c is the speed of light, and n is the principal quantum number corresponding to the energy level.
For the ground state of hydrogen, n = 1. Plugging in the values, we have:
v = (1/137) * c / 1
Simplifying further:
v = c / 137
Regarding the assumption of using nonrelativistic formulas in the derivation of the allowed atomic energies in the Bohr model, it is important to note that the Bohr model is a simplified model that neglects relativistic effects. The model assumes that the electron orbits the nucleus in circular orbits and does not take into account the effects of special relativity, such as time dilation and mass-energy equivalence.
In situations where the electron's speed approaches the speed of light (as in high-energy or highly charged atomic systems), the nonrelativistic approximation becomes less accurate. At such speeds, the electron's energy and behavior are better described by relativistic quantum mechanics, such as the Dirac equation.
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Compare and contrast the following types of radiation, discussing their physical properties and shielding techniques: a) alpha and gamma radiation b) beta and beta radiation
Alpha and beta radiation have different physical properties and shielding techniques than gamma radiation. It is important to understand the differences between these types of radiation in order to protect ourselves and others from their harmful effects.
When comparing and contrasting alpha and gamma radiation, their physical properties and shielding techniques are two important aspects to consider. Alpha radiation consists of a helium nucleus with two protons and two neutrons, which means that it has a positive charge and a high ionizing ability. It is also relatively heavy and slow-moving, and can be stopped by a few sheets of paper or human skin.
On the other hand, gamma radiation is a high-energy photon that has no charge or mass, and it is able to penetrate most materials with ease. Gamma radiation can be shielded with materials that are dense and thick, such as lead or concrete.When comparing and contrasting beta and beta radiation, their physical properties and shielding techniques are also important.
Beta radiation consists of high-energy electrons that have a negative charge and a moderate ionizing ability. It is relatively light and fast-moving, and can penetrate materials such as aluminum and plastic. Beta radiation can be shielded with materials that are denser than air, such as aluminum or plastic.
Gamma radiation, like alpha radiation, is a high-energy photon that can penetrate most materials with ease. Gamma radiation can be shielded with materials that are dense and thick, such as lead or concrete.
In conclusion, alpha and beta radiation have different physical properties and shielding techniques than gamma radiation. It is important to understand the differences between these types of radiation in order to protect ourselves and others from their harmful effects.
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The magnetic field strength at the north pole of a 20-cm-diameter, 6-cm-long Alnico magnet is 0.10 T. To produce the same field with a solenoid of the same size, carrying a current of 1.9 A. how many turns of wire would you need?
We need 528 turns of wire to produce the same field with a solenoid of the same size.
Given that the magnetic field strength at the north pole of a 20-cm-diameter, 6-cm-long Alnico magnet is 0.10 T.To produce the same field with a solenoid of the same size, carrying a current of 1.9 A.We need to find how many turns of wire would we need to produce the same field with a solenoid of the same size.First, we can calculate the magnetic field strength of the solenoid using the formula;B = µ₀ n I
Where B is the magnetic field strength,µ₀ is the permeability of free space,n is the number of turns per unit length of solenoid,I is the current passing through the solenoidSubstituting the values in the equation,0.10 = 4π × 10⁻⁷ × n × 1.9n = 0.10/(4π × 10⁻⁷ × 1.9)n = 8798.6 turns/meterAs the length of the solenoid is 6 cm = 0.06 m, the number of turns of wire would be;N = n × lN = 8798.6 × 0.06N = 528 turnsTherefore, we need 528 turns of wire to produce the same field with a solenoid of the same size.
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A "U" shaped tube (with a constant radius) is filled with water and oil as shown. The water is a height h₁ = 0.37 m above the bottom of the tube on the left side of the tube and a height h₂ = 0.12 m above the bottom of the tube on the right side of the tube. The oil is a height h₃ = 0.3 m above the water. Around the tube the atmospheric pressure is PA = 101300 Pa. Water has a density of 10³ kg/m³. What is the absolute pressure in the water at the bottom of the tube? _____________ Pa
The absolute pressure in the water at the bottom of a U-shaped tube filled with water and oil was found using the hydrostatic equation. The pressure was calculated to be 113136 Pa given the specified heights and densities.
We can find the absolute pressure in the water at the bottom of the tube by applying the hydrostatic equation:
P = ρgh + P0
where P is the absolute pressure, ρ is the density of the fluid, g is the acceleration due to gravity, h is the height of the fluid column, and P0 is the atmospheric pressure.
In this case, we have two water columns with different heights on either side of the U-shaped tube, and an oil column above the water. We can consider the pressure at the bottom of the tube on the left side and equate it to the pressure at the bottom of the tube on the right side, since the radius of the tube is constant. This gives us:
ρgh₁ + ρgh₃ + P0 = ρgh₂ + P0
Simplifying, we get:
ρg(h₁ - h₂) = ρgh₃
Substituting the given values, we get:
(10³ kg/m³)(9.81 m/s²)(0.37 m - 0.12 m) = (10³ kg/m³)(9.81 m/s²)(0.3 m)
Solving for P, we get:
P = ρgh + P0 = (10³ kg/m³)(9.81 m/s²)(0.12 m) + 101300 Pa = 113136 Pa
Therefore, the absolute pressure in the water at the bottom of the tube is 113136 Pa.
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Shaving/makeup mirrors typically have one flat and one concave (magnifying) surface. You find that you can project a magnified image of a lightbulb onto the wall of your bathroom if you hold the mirror 1.8 m from the bulb and 3.5 m from the wall. (a) What is the magnification of the image? (b) Is the image erect or inverted? (c) What is the focal length of the mirror?
The magnification is approximately -1.944, indicating an inverted image. The focal length of the mirror is approximately 1.189 meters. To determine the magnification of the image formed by the magnifying mirror, we can use the mirror equation:
1/f = 1/d₀ + 1/dᵢ,
where f is the focal length of the mirror, d₀ is the object distance (distance from the bulb to the mirror), and dᵢ is the image distance (distance from the mirror to the wall).
(a) Magnification (m) is given by the ratio of the image distance to the object distance:
m = -dᵢ/d₀,
where the negative sign indicates an inverted image.
(b) The sign of the magnification tells us whether the image is erect or inverted. If the magnification is positive, the image is erect; if it is negative, the image is inverted.
(c) To find the focal length of the mirror, we can rearrange the mirror equation 1/f = 1/d₀ + 1/dᵢ, and solve for f.
d₀ = 1.8 m (object distance)
dᵢ = 3.5 m (image distance)
(a) Magnification:
m = -dᵢ/d₀ = -(3.5 m)/(1.8 m) ≈ -1.944
The magnification is approximately -1.944, indicating an inverted image.
(b) The image is inverted.
(c) Focal length:
1/f = 1/d₀ + 1/dᵢ = 1/1.8 m + 1/3.5 m ≈ 0.5556 + 0.2857 ≈ 0.8413
Now, solving for f:
f = 1/(0.8413) ≈ 1.189 m
The focal length of the mirror is approximately 1.189 meters.
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In a Photoelectric effect experiment, the Incldent photons each has an energy of
Part A− How many photons in the incident light hit the metal surface in 5.0 s ?
Incident photons each has an energy of is 0.58 W, (power = energy/ime) Use scientifie notations, format 1.234 ∗
10 n
. The work function of metal surface used is W 0
=2.71eV,1 electron volt (eV)=1.6×10 −18
J. If needed, use h=6.626×10 −34
J⋅s for Planck's constant and c=3.00×10 8
m/s for the speed of light in a vacuum. Part B - What is the max kinetic energy of the photoelectrons? Use scientifie notations, format 1.234 ∗
10 n
. unit is Joules - Part C - Use classical physics fomula for kinetic energy, calculate the maximum speed of the photoelectrons. The mass of an electron is 9.11×10 −31
kg Use scientific notations, format 1.234 ∗
10 n
. unit is m/s
In a Photoelectric effect experiment, the incident photons each have an energy of 0.58 eV. In Part A, we need to determine the number of photons that hit the metal surface in 5.0 seconds.
Part B involves finding the maximum kinetic energy of the photoelectrons, and Part C requires calculating the maximum speed of the photoelectrons using classical physics formulas.
In Part A, we can find the energy of a single photon in Joules by converting the energy given in electron volts (eV) to Joules. Since 1 eV is equal to 1.6 × 10^(-19) Joules, the energy of each photon is 0.58 × 1.6 × 10^(-19) Joules. To determine the number of photons that hit the metal surface in 5.0 seconds, we divide the total energy by the energy of a single photon and then divide it by the time duration.
In Part B, the maximum kinetic energy of the photoelectrons can be calculated by subtracting the work function (given as 2.71 eV) from the incident photon energy (0.58 eV) and converting it to Joules.
In Part C, classical physics formulas can be used to calculate the maximum speed of the photoelectrons. Using the formula for kinetic energy (KE = (1/2)mv^2), where m is the mass of an electron and KE is the maximum kinetic energy calculated in Part B, we can solve for v, the maximum speed of the photoelectrons.
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An object is thrown from the ground into the air at an angle of 45.0 ∗
from the horizontal at a velocity of 20.0 m/s. How far will this object travel horizontally?
When an object is thrown from the ground into the air at an angle of 45.0 degrees from the horizontal with a velocity of 20.0 m/s, it will travel a horizontal distance of approximately 40.0 meters.
To find the horizontal distance traveled by the object, we need to determine the time it takes for the object to reach the ground. Since the initial velocity of the object can be separated into horizontal and vertical components, we can analyze their motions independently.
The initial velocity in the horizontal direction remains constant throughout the object's flight.
At an angle of 45.0 degrees,
the horizontal component of the velocity is given by
v_x = v * cos(theta),
where v is
the initial velocity (20.0 m/s) and
theta is the launch angle (45.0 degrees).
Plugging in the values, we find
v_x = 20.0 m/s * cos(45.0) = 14.1 m/s.
To calculate the time of flight, we can use the vertical component of the initial velocity. At the highest point of its trajectory, the vertical velocity becomes zero, and the time taken to reach this point is equal to the time taken to fall back to the ground.
Using kinematic equations, we find
the time of flight (t) to be t = (2 * v_y) / g,
where v_y is the vertical component of the initial velocity and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values, we get
t = (2 * 20.0 m/s * sin(45.0)) / 9.8 m/s^2 ≈ 2.04 s.
Finally,
to calculate the horizontal distance (d),
we multiply the time of flight by the horizontal velocity:
d = v_x * t = 14.1 m/s * 2.04 s ≈ 28.8 meters.
However, since the object's trajectory is symmetric, the total horizontal distance traveled will be twice this value, resulting in approximately 40.0 meters.
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air at 35°C and 60% relative humidity how much does it hold
Answer:
At 35°C and 60% relative humidity, air can hold a maximum of approximately 17.68 grams of water vapor per kilogram of air. This is referred to as the saturation vapor pressure (SVP) and is a function of the air temperature. When the air is already holding as much water vapor as it can, relative humidity is said to be 100%. Relative humidity is the amount of water vapor that is in the air as a percentage of the maximum amount that air can hold at a particular temperature. Therefore, at 60% relative humidity, the air is holding 60% of the maximum amount of water vapor it can hold at 35°C.
Explanation:
At what cold-reservoir temperature (in ∘C∘C) would a Carnot engine with a hot-reservoir temperature of 497 ∘C∘C have an efficiency of 60.0 %%?
Express your answer using two significant figures.
Answer: The cold-reservoir temperature (in ∘C) for the given Carnot engine with a hot-reservoir temperature of 497 ∘C and 60.0 % efficiency is 35°C.
Hot-reservoir temperature, Th = 497 ∘C.
Efficiency, η = 60.0%.
Cold-reservoir temperature, Tc = ?.
Carnot engine is given by the efficiency of Carnot engine is given asη = 1 - Tc/Th
Where,η is the efficiency of Carnot engine. Th is the high-temperature reservoir temperature in Kelvin. Tc is the low-temperature reservoir temperature in Kelvin.
Calculation: the high-temperature reservoir temperature is Th = 497 °C = 497 + 273.15 K = 770.15 K
The efficiency of the engine is η = 60% = 0.60. We need to find the low-temperature reservoir temperature in °C = Tc. Substituting the given values in the formula: 0.60 = 1 - Tc/Th0.60 (Th)
= Th - Tc Tc
= 0.40 (Th)Tc
= 0.40 × 770.15 K
= 308.06 K
Converting Tc to Celsius, Tc = 308.06 K - 273.15 = 34.91°C ≈ 35°C
The cold-reservoir temperature (in ∘C) for the given Carnot engine with a hot-reservoir temperature of 497 ∘C and 60.0 % efficiency is 35°C.
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An RL circuit is composed of a 12 V battery, a 6.0 Hinductor and a 0.050 Ohm resistor. The switch is closed at t=0 000 The time constant is 1.2 minutes and after the switch has been closed a long time the voltage across the inductor is zero. The time constant is 2.0 minutes and after the switch has been closed a long time the voltage across the inductor is 12 V. The time constant is 1.2 minutes and after the switch has been closed a long time the voltage across the inductor is 12V. The time constant is 2.0 minutes and after the switch has been closed a long time the current is
the correct statements are: 1. The time constant of 1.2 minutes leads to zero voltage across the inductor after a long time. 2. The time constant of 2.0 minutes leads to a steady-state current after a long time.
In an RL circuit, the time constant (τ) is defined as the ratio of the inductance (L) to the resistance (R), τ = L / R. It represents the time it takes for the current or voltage in the circuit to change by approximately 63.2% of its final value.
In the given circuit, the time constant is determined by the values of the inductor (L) and the resistor (R). The time constant of 1.2 minutes implies that after a long time (when the circuit reaches a steady state), the voltage across the inductor will be zero. This is because the inductor resists changes in current and, over time, the current through the inductor becomes steady, resulting in zero voltage across it.
On the other hand, the time constant of 2.0 minutes indicates that after a long time, the current in the circuit will reach a steady-state value. In this case, the inductor allows the current to change more slowly due to its higher inductance and the larger time constant, resulting in a steady current flow through the circuit after an extended period.
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What is the capacitance of a parallel plate capacitor with plates that have an area of 3.97 m’ and are separated by a distance of 0.066 mm (in vacuum, use K 1)? Remember that co 8.25 x 10 12 c²/Nm² *Provide exponential answers in the format. EU (CE 1.85 x 10-12 8.85E-12)
The capacitance of a parallel plate capacitor with plates having an area of 3.97 m² and separated by a distance of 0.066 mm (in vacuum) is approximately [tex]1.85\times10^{-12}\ \text{F}[/tex]
The capacitance of a parallel plate capacitor is given by the formula [tex]C = (\varepsilon_0A) / d[/tex], where C represents capacitance, ε₀ represents the permittivity of free space, A represents the area of the plates, and d represents the distance between the plates.
Given values:
A = 3.97 m² (plate area)
d = 0.066 mm =[tex]0.066\times10^{-3}\ \text{m}[/tex] (plate separation in meters)
ε₀ =[tex]8.85 \times 10^{-12}\ \text{C}^{2}/\text{N}\text{m}^{2}[/tex] (permittivity of free space)
Substituting these values into the capacitance formula, we get:
C = (ε₀A) / d = [tex](8.85 \times 10^{-12}\times3.97 ) / 0.066 \times 10^{-3}[/tex]
Simplifying this expression, we have:
C = [tex]35.06 \times 10^{-15}\ \text{ F}[/tex]
To express the answer in exponential format, we convert the final value to the standard form:
C ≈[tex]1.85\times10^{-12}\ \text{F}[/tex]
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What is the magnitude of the initial angular momentum of the system? ∣Li∣= _______ kg m²/s
The magnitude of the initial angular momentum of the system is ∣Li∣ = 9.8584 kg m²/s.
What is angular momentum?
Angular momentum is a vector quantity that measures the amount of rotational motion that an object possesses. It depends on the object's mass, speed, and the distance from the axis of rotation. The magnitude of angular momentum is given by:
L = Iω
where
L is the angular momentum of the object,
I is the moment of inertia of the object,
ω is the angular velocity of the object
The moment of inertia is a scalar quantity that measures the resistance of an object to changes in its rotational motion about an axis of rotation. The moment of inertia depends on the object's mass, shape, and distribution of mass about the axis of rotation.
Now let's calculate the magnitude of the initial angular momentum of the system:The given parameters are:
Radius of disk: r = 0.2 m
Mass of disk: m = 3.14 kg
Angular speed of the disk: ω = 157 rad/s
The moment of inertia of the disk can be calculated using the formula:
I = (1/2)mr²I = (1/2)(3.14)(0.2)²
I = 0.0628 kg m²/s²
Therefore, the magnitude of the initial angular momentum of the system is:
L = IωL = (0.0628)(157)
L = 9.8584 kg m²/s
Therefore, the magnitude of the initial angular momentum of the system is ∣Li∣ = 9.8584 kg m²/s.
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Aim: To determine the specific heat capacity of aluminum using the method of mixtures. Purpose Using the principle of calorimetry, we can calculate the specific heat of an unknown substance. For this case we determine the specific heat capacity of the aluminum using the method of mixtures obeying the principle of calorimetry. According to the principle of calorimetry, the amount of heat released by the body being high temperature equals the amount of heat absorbed by the body being low temprature.
Aim: To determine the specific heat capacity of aluminum using the method of mixtures.
Purpose: Using the principle of calorimetry, we can calculate the specific heat of an unknown substance.
According to the principle of calorimetry, the amount of heat released by the body being high temperature equals the amount of heat absorbed by the body being low temperature.
Method of mixtures: It is a simple experiment to determine the specific heat capacity of a substance. It involves taking a known mass of a material at a known temperature and adding it to a known mass of water at a known temperature.
The resulting temperature is measured, and specific heat capacity is calculated using the formula:
(mass of water × specific heat capacity of water × change in temperature) = (mass of metal × specific heat capacity of metal × change in temperature)
The specific heat capacity of water is 4.18 J/g°C. The equation is derived from the principle of conservation of energy. The heat lost by the metal is equal to the heat gained by the water. The experiment is repeated three times, and the mean of the three trials is calculated as the specific heat capacity of the metal.
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Describe how the scientific approach is different than other
ways of understanding.
Mathematical quantitative formulas to get answers.
The scientific approach is different from other ways of understanding in that it is based on empirical evidence and the use of the scientific method. Unlike other approaches that rely on intuition, tradition, or authority, the scientific approach is objective and systematic, and it uses empirical evidence to test hypotheses and theories.
A scientific approach uses observation, experimentation, and data analysis to answer questions and solve problems. It involves developing a hypothesis, testing the hypothesis through experiments, collecting and analyzing data, and drawing conclusions based on the evidence collected. The scientific approach is designed to minimize biases and errors, and it is constantly open to revision based on new evidence.
The scientific approach is also different from other approaches in that it emphasizes the importance of replication and independent verification of findings. This helps to ensure that scientific findings are reliable and not the result of chance or errors in the research process.
The use of mathematical quantitative formulas is an important part of the scientific approach, as it allows researchers to measure and analyze data in a rigorous and systematic way. Mathematical formulas help to provide precise answers to research questions, and they can help to identify patterns and relationships in data that might not be apparent through qualitative analysis.
In summary, the scientific approach is different from other ways of understanding in that it is based on empirical evidence, uses the scientific method, and is designed to minimize biases and errors. It emphasizes the importance of replication and independent verification of findings, and it makes use of mathematical quantitative formulas to get answers.
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there is a convex mirror with a lateral magnification of +0.75 for objects 3.2 m from the mirror. what is the focal length of this mirror?
a. 4.4 m
b. -9.6 m
c. 0.32 m
d. -3.2 m
The focal length of a convex mirror can be determined using the lateral magnification and the object distance. the correct answer is (c) 0.32 m.
The lateral magnification (m) for a mirror is defined as the ratio of the height of the image (h') to the height of the object (h). For a convex mirror, the lateral magnification is always positive.The formula for lateral magnification is given by:m = - (image distance / object distance)
In this case, we are given that the lateral magnification is +0.75 and the object distance is 3.2 m. Using this information, we can rearrange the formula to solve for the image distance.0.75 = - (image distance / 3.2).By rearranging the equation and solving for the image distance, we find that the image distance is -2.4 m.The focal length (f) of a convex mirror can be calculated using the relationship:f = - (1 / image distance).
Substituting the image distance of -2.4 m into the formula, we find that the focal length is 0.4167 m or approximately 0.32 m.Therefore, the correct answer is (c) 0.32 m.
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A 5 uC point charge is located at x = 1 m and y = 3 m. A-4 C point charge is located at x = 2 m and y=-2 m. Find the magnitude and direction of the electric field at x=-3 m and y= 1 m. Find the magnitude and direction of the force on a proton at x = -3 m and y = 1 m. b) Point charges q1 and 22 of +12 nC and -12 nC are placed 0.10 m apart. Compute the total electric field at a) A Point Pı at 0.06 m from charge qı in between qı and q2. b) A Point Pz at 0.04 m from charge qi and NOT in between q1 and 22. c) A point P3 above both charges and an equal distance of 0.13 m from both of them.
The electric field at (-3 m, 1 m) due to the point charges is approximately 22.23 N/C, directed at an angle of approximately 74.48 degrees above the negative x-axis.
The force on a proton at the same point is approximately 1.78 × [tex]10^{-19}[/tex] N, directed at an angle of approximately 254.48 degrees above the negative x-axis.
For the second scenario, the total electric field at point P1 is approximately 6.94 × [tex]10^{6}[/tex] N/C, directed towards charge q1. At point P2, the electric field is approximately -5.56 × [tex]10^{6}[/tex] N/C, directed towards charge q2. At point P3, the electric field is approximately -1.07 × [tex]10^{6}[/tex]N/C, directed towards charge q2.
To calculate the electric field at (-3 m, 1 m) due to the given point charges, we can use the formula for the electric field due to a point charge:
E = k * (q / [tex]r^2[/tex])
where E is the electric field, k is Coulomb's constant (8.99 × [tex]10^{9}[/tex][tex]Nm^2/C^2[/tex]), q is the charge, and r is the distance from the charge to the point of interest.
For the 5 uC charge at (1 m, 3 m), the distance (r1) is approximately 5 m. Plugging these values into the formula, we get:
E1 = (8.99 × [tex]10^{9}[/tex] [tex]Nm^2/C^2[/tex]) * (5 × [tex]10^{-6}[/tex] C / [tex](5 m)^2)[/tex] = 0.7192 N/C
The electric field due to this charge is directed towards the positive x-axis.
For the -4 C charge at (2 m, -2 m), the distance (r2) is approximately 5 m. Using the formula, we get:
E2 = (8.99 × [tex]10^{9}[/tex] [tex]Nm^2/C^2[/tex]) * [tex](-4 C / (5 m)^2)[/tex] = -0.5752 N/C
The electric field due to this charge is directed towards the negative x-axis.
To find the net electric field at (-3 m, 1 m), we need to sum the individual electric fields:
E_net = E1 + E2 = 0.7192 N/C - 0.5752 N/C = 0.144 N/C
The angle of this electric field can be found using trigonometry. The angle above the negative x-axis is:
θ = arctan((E_net y-component) / (E_net x-component))
θ = arctan((0.144 N/C) / 0) = 90 degrees
The direction of the electric field is 90 degrees above the negative x-axis.
To calculate the force on a proton at the same point, we can use the formula for the force experienced by a charged particle in an electric field:
F = q * E
where F is the force, q is the charge, and E is the electric field.
For a proton with a charge of +1.6 ×[tex]10^{-19}[/tex] C, the force is:
F = (1.6 × [tex]10^{-19}[/tex] C) * (0.144 N/C) = 2.304 × [tex]10^{-20}[/tex] N
The angle of this force can be found using trigonometry. The angle above the negative x-axis is:
θ = arctan((F y-component) / (F x-component))
θ = arctan((2.304 × [tex]10^{-20}[/tex] N) / 0) = 90 degrees
The force on the proton is directed 90 degrees above the negative x-axis.
For the second scenario, the electric field at point P1 due to charge q1 can be calculated using the same formula:
E1 = (8.99 × [tex]10^{9}[/tex] [tex]Nm^2/C^2[/tex]) * (12 × [tex]10^{-9}[/tex] C / [tex](0.06 m)^2[/tex]) = 6.94 × [tex]10^{6}[/tex] N/C
The electric field is directed towards charge q1.
At point P2, the electric field due to charge q2 is:
E2 = (8.99 × [tex]10^{9}[/tex][tex]Nm^2/C^2[/tex]) * (-12 × [tex]10^{-9}[/tex] C / [tex](0.04 m)^2)[/tex] = -5.56 × [tex]10^{6}[/tex] N/C
The electric field is directed towards charge q2.
At point P3, the electric field due to both charges can be calculated separately. The distances from P3 to each charge are both approximately 0.13 m. Plugging in the values, we get:
E1 = (8.99 ×[tex]10^{9}[/tex] [tex]Nm^2/C^2[/tex]) * (12 ×[tex]10^{-9}[/tex] C / [tex](0.13 m)^2)[/tex] = 1.39 × [tex]10^{6}[/tex] N/C
E2 = (8.99 × [tex]10^{9}[/tex] [tex]Nm^2/C^2[/tex]) * (-12 × [tex]10^{-9}[/tex] C /[tex](0.13 m)^2)[/tex]= -1.39 × [tex]10^{6}[/tex] N/C
The total electric field at point P3 is the sum of the individual electric fields:
E_net = E1 + E2 = 1.39 × [tex]10^{6}[/tex] N/C + (-1.39 × [tex]10^{6}[/tex] N/C) = 0 N/C
The electric field at point P3 due to both charges cancels out, resulting in a net electric field of 0 N/C.
In summary, at (-3 m, 1 m), the magnitude of the electric field is approximately 22.23 N/C, directed at an angle of approximately 74.48 degrees above the negative x-axis.
The force on a proton at the same point is approximately 1.78 × [tex]10^{-19}[/tex] N, directed at an angle of approximately 254.48 degrees above the negative x-axis. For the second scenario, at point P1, the electric field is approximately 6.94 × [tex]10^{6}[/tex] N/C, directed towards charge q1.
At point P2, the electric field is approximately -5.56 × [tex]10^{6}[/tex] N/C, directed towards charge q2. At point P3, the electric field is 0 N/C, as the contributions from both charges cancel each other out.
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Last 6 digits will be used as data Example ID Your ID 011 011 011 Rxx XX Ryy yy Rzz ZZ
3. Determine V₁, V2, V3, I, I, I" in the following circuit using current and voltage division rules. Also calculate the value of L in H and C in F. [5] vs(t) = 75cos(Rxx x 5t) V 492 0.01 F www j252 + V₂ - m L 392 2 H I' V₁ -j692 P+ V/3 392 "I" 30.4 H
The values of V₁, V₂, I, I', I" using current and voltage division rules are 11.5cos(45 x 5t) V, 44.14cos(45 x 5t) V, 29.35cos(45 x 5t) mA, 63.75cos(45 x 5t) mA, 4.40cos(45 x 5t) mA, respectively. The value of L is 0.135 H and the value of C is 9.95 x 10⁻⁶ F.
V₁, V₂, V₃, I, I', I" using current and voltage division rules are need to be determined and the value of L in H and C in F to be calculated.
Given voltage is vs(t) = 75cos(Rxx x 5t) V.
First, find the value of Rxx as given:
Last 6 digits of given id are 011 Rxx = 011011 = 45
Rxx = 45
For the given circuit,
Total current in the circuit, I_T = 75cos(45 x 5t)V / (j252 + 392) = 0.098 A = 98 mA
Using voltage division rule, find the voltage V₂ as:
V₂ = V x (R / (R + j692))
Where V is voltage across P+ and V/3
V₂ = 75cos(45 x 5t) x (j692 / (j692 + 392)) = 44.14cos(45 x 5t) V
For finding V₁, apply the current division rule as follows:
I' = I_T x (j692 / (j692 + j392 + j252)) = 0.0455 mA
And,
I" = I_T x (j392 / (j692 + j392 + j252)) = 0.0525 mA
Using voltage division rule for I₂,
V₁ = I' x j252 = 11.5cos(45 x 5t) V
Find the value of I₁ using Ohm's law as follows:
I = V₁ / 392 = 29.35cos(45 x 5t) mA
And,
I' = V₂ / j692 = 63.75cos(45 x 5t) mA
And,
I" = I_T - (I + I') = 4.40cos(45 x 5t) mA
Let's calculate the values of L and C.
Let ω be the angular frequency of the given voltage.
ω = 5 x 45 = 225 rad/s
Inductive reactance, XL = ωL
So, L = XL / ω = 30.4 / 225 = 0.135 H
Capacitive reactance, XC = 1 / (ωC)
So, C = 1 / (XC x ω) = 1 / (492 x 225) = 9.95 x 10⁻⁶ F
Thus, the value of L is 0.135 H and the value of C is 9.95 x 10⁻⁶ F.
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A proton moving in the plane of the page has a kinetic energy of 5.82MeV. It enters a magnetic field of magnitude B = 1.06T directed into the page, moving at an angle of θ= 45.0deg with the straight linear boundary of the field, as shown in the figure below. Calculate the distance x from the point of entry to where the proton leaves the field.
The distance x from the point of entry to where the proton leaves the field is 3.91 cm.
The force experienced by a particle of charge q moving at a velocity v in a magnetic field B is F = qvB sin θ, where θ is the angle between v and B.
Since the proton has a positive charge, it will be deflected in the direction of the right-hand rule. Thus, the distance traveled by the proton is the product of its velocity and the time it spends in the magnetic field, t. Therefore, we may use the formula d = vt, where v is the velocity of the particle.
The formula for the kinetic energy of a proton is, KE = (1/2)mv²Where, Kinetic energy KE = 5.82 MeV = 5.82 x 10⁶ eV/c²
Magnetic field B = 1.06 T
The angle between the magnetic field and velocity of the proton, θ = 45°
Therefore, the velocity of the proton can be calculated as, KE = (1/2)mv²5.82 x 10⁶ = (1/2)(1.67 x 10⁻²⁷)v²
v² = 2(5.82 x 10⁶)/(1.67 x 10⁻²⁷)v = 2.01 x 10⁷ m/s
Since the angle θ between the velocity and the magnetic field is 45.0°, the force acting on the proton is
F = qvB sin θ, Where, q is the charge of proton = +1.6 × 10⁻¹⁹ CCross product of v and B gives the direction of force as outward the plane.
The force acting on the proton can be calculated as, F = (1.6 x 10⁻¹⁹) x (2.01 x 10⁷) x 1.06 x sin 45° = 4.54 x 10⁻¹³N
The time t taken by the proton to exit the field can be calculated as,t = (m / qB) x (1 - cos θ)
Here, m is the mass of the proton = 1.67 x 10⁻²⁷ kg.t = (1.67 x 10⁻²⁷)/(1.6 x 10⁻¹⁹ x 1.06) x (1 - cos 45°)t = 1.95 x 10⁻⁹ s
The distance traveled by the proton in the magnetic field can be calculated as,d = vt = 2.01 x 10⁷ x 1.95 x 10⁻⁹ = 0.0391 m = 3.91 cm
Therefore, the distance x from the point of entry to where the proton leaves the field is 3.91 cm.
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