Given that the object is placed 24.85 cm in front of a diverging lens which has a focal length with a magnitude of 11.52 cm. Let the distance of the image formed be v, and the distance of the object be u.
Using the lens formula, 1/f = 1/v − 1/u. Since it's a diverging lens, the focal length is negative, f = -11.52 cm, Plugging the values, we have;1/(-11.52) = 1/v − 1/24.85 cm, solving for v; v = -13.39 cm or -0.1339 m. Since the image is larger than we want, it means the image formed is virtual, erect, and magnified.
The magnification is given by; M = -v/u. From the formula above, we have; M = -(-0.1339)/24.85M = 0.0054The negative sign in the magnification indicates that the image formed is virtual and erect, which we have already stated above. Also, the magnification value indicates that the image formed is larger than the object.
In order to produce an image that is reduced by a factor of 3.8, we can use the magnification formula; M = -v/u = −3.8.By substitution, we have;-0.1339/u = −3.8u = -0.1339/(-3.8)u = 0.03521 m = 3.52 cm.
Therefore, the distance of the object should be placed 3.52 cm in front of the lens in order to produce an image that is reduced by a factor of 3.8.
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(a) What is room temperature (68°F) in
°C and K? (b) What
is the boiling temperature of liquid nitrogen (77 K) in °C and °F?
Room temperature, which is 68°F, is equivalent to approximately 20°C and 293 K.
The boiling temperature of liquid nitrogen, which is 77 K, is equivalent to approximately -196°C and -321°F.
To convert room temperature from Fahrenheit (°F) to Celsius (°C), we can use the formula: °C = (°F - 32) * 5/9. Substituting 68°F into the formula, we get: °C = (68 - 32) * 5/9 ≈ 20°C.
To convert from Celsius to Kelvin (K), we simply add 273.15 to the Celsius value. Therefore, 20°C + 273.15 ≈ 293 K.
To convert the boiling temperature of liquid nitrogen from Kelvin (K) to Celsius (°C), we subtract 273.15. Therefore, 77 K - 273.15 ≈ -196°C.
To convert from Celsius to Fahrenheit, we can use the formula: °F = (°C * 9/5) + 32. Substituting -196°C into the formula, we get: °F = (-196 * 9/5) + 32 ≈ -321°F.
Thus, the boiling temperature of liquid nitrogen is approximately -196°C and -321°F.
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A sinusoidal sound wave moves through a medium and is described by the displacement wave function
(x, t) = 2.19 cos(16.3x - 851t)
where s is in micrometers, x is in meters, and t is in seconds.
(a) Find the amplitude of this wave.
um
(b) Find the wavelength of this wave.
cm
(c) Find the speed of this wave.
(a) The amplitude of the sinusoidal sound wave is 2.19 μm.
(b) The wavelength is given by λ = 1/16.3 = 0.0613 m or 6.13 cm.
(c) The frequency is f = 851 Hz. S
The amplitude of a wave represents the maximum displacement of particles in the medium from their equilibrium position. In this case, the maximum displacement is given as 2.19 μm. Moving on to the wavelength, it can be determined by examining the coefficient of x in the displacement wave function, which is 16.3.
This coefficient represents the number of wavelengths that fit within a distance of 1 meter. Therefore, the wavelength is calculated as 1/16.3 = 0.0613 m or 6.13 cm. To find the speed of the wave, the formula v = λf is used, where v is the speed, λ is the wavelength, and f is the frequency. The frequency is obtained from the coefficient of t in the displacement wave function, which is 851. Substituting the values, the speed is calculated as (0.0613 m) × (851 Hz) = 52.15 m/s.
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3-A ball is dropped from the top of a tall building. Assuming free fall, how far does the ball fall in 1.50 s?
1-A 1kg ball is fired from a cannon. What is the change in the ball’s kinetic energy when it accelerates form 4.0 m/s2 to 8 m/s2?
Therefore, the change in kinetic energy of the ball when it accelerates from 4.0 m/s² to 8 m/s² is 24 J.
3-A ball is dropped from the top of a tall building. Assuming free fall, how far does the ball fall in 1.50 s?
For a body in free fall, the distance (d) traveled can be calculated using the formula:
d = (1/2)gt²
Where g = 9.8 m/s² is the acceleration due to gravity and t is the time taken.
Therefore, using the given values, we have:
d = (1/2)gt²d = (1/2)(9.8 m/s²)(1.50 s)²
d = 17.6 m
Therefore, the ball falls a distance of 17.6 m in 1.50 s assuming free fall.
1-A 1kg ball is fired from a cannon.
What is the change in the ball’s kinetic energy when it accelerates form 4.0 m/s² to 8 m/s²?
The change in kinetic energy (ΔK) of a body is given by the formula:
ΔK = (1/2) m (v₂² - v₁²)
Where m is the mass of the body, v₁ is the initial velocity, and v₂ is the final velocity.
Therefore, using the given values, we have:
ΔK = (1/2) (1 kg) [(8 m/s)² - (4 m/s)²]
ΔK = (1/2) (1 kg) [64 m²/s² - 16 m²/s²]
ΔK = (1/2) (1 kg) (48 m²/s²)
ΔK = 24 J
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Q. 137: Two lenses L₁ and L₂ are used to make a telescope. The larger lens L₁ is a convex lens with both surfaces having radius of curvature equal to 0.5 m. The smaller lens L₂ has two surfaces with radius of curvature 4 cm. Both the lenses are made of glass having refractive index 1.5. The two lenses are mounted in a tube with separation between them equal to 1 cm less than the sum of their focal length. (a) Find the position of the image formed by such a telescope for an object at a distance of 100 m from the objective lens L₁. (b) What is the size of the image if object is 1 m high? Do you think that lateral magnification is a useful way to characterize a telescope?
a) The image is 6.74 times larger than the object and is formed 6.74 times farther from the objective lens than the focal length.
b) The image is 6.74 times larger than the object and is formed 6.74 times farther from the objective lens than the focal length.
(a) Position of the image formed by such a telescope for an object at a distance of 100m from the objective lens L₁
The focal length of the convex lens L₁ can be obtained as follows:f = R/(n-1)
where R is the radius of curvature of the lens and n is the refractive index.
f = 0.5 m / (1.5 - 1) = 1 m
The distance between the two lenses is given as 1 cm less than the sum of their focal length. The focal length of the smaller lens L₂ is given as:
f₂ = R/(n-1) = 0.04m/(1.5-1) = 0.16 m
The distance between the lenses is given as (f₁ + f₂ - 0.01) = 1 + 0.16 - 0.01 = 1.15 m
Therefore, the magnification of the telescope is given by:
M = - v/u
where v is the image distance and u is the object distance.
u = -100 m, f₁ = 1 m, and f₂ = 0.16 m
Substituting in the formula,
M = - (f₁ + f₂ - d)/(f₂ * (f₁ + f₂ - d)/f₁ - d/u)
M = - (1.16 - 0.01)/((0.16 * (1.16 - 0.01))/1 - (-100)) = -6.74
We obtain a negative magnification because the image is inverted.
(b) Size of the image if object is 1m high
The height of the image is given by:
h₂ = M * h₁
where h₁ is the height of the objecth₁ = 1 m
Therefore, the height of the image is:
h₂ = -6.74 * 1 = -6.74 m
We obtain a negative height because the image is inverted.
Lateral magnification is a useful way to characterize a telescope as it provides information about the size and position of the image relative to the object. It helps to understand the quality of the image and how well the telescope is able to resolve details.
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A hair dryer and a curling iron have resistances of 15 Q2 and 25 Q2, respectively, and are connected in series. They are connected to a 60 V battery. Calculate the power used by the hair dryer. A hair dryer and a curling iron have resistances of 15 2 and 25 2, respectively, and are connected in series. They are connected to a 60 V battery. Calculate the power used by the curling iron.
The power used by the hair dryer is 240 watts. To calculate the power used by each appliance, we need to use the formulas for power and resistance. The power formula is:
P = V^2 / R:
P is the power in watts (W)
V is the voltage in volts (V)
R is the resistance in ohms (Ω)
Resistance of the hair dryer, R_hairdryer = 15 Ω
Voltage across the hair dryer, V_hairdryer = 60 V
P_hairdryer = V_hairdryer^2 / R_hairdryer
= (60 V)^2 / 15 Ω
= 3600 V^2 / 15 Ω
= 240 W
Therefore, the power used by the hair dryer is 240 watts.
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Complete the following beta decays. Make sure to delete the "0" that appears in the pre-subscript position of the nuclide symbol before submitting your responses. 20F e +ve+ 239Npe" + vet 3H+ve+ 207 Bi et + 60 e + ve + Ni
Please note that the "vet" in the second decay is not a recognized symbol or notation for beta decay. If you provide more specific information or correct any errors.
Neutrinos are subatomic particles that are electrically neutral and have very low mass. They interact weakly with matter, making them difficult to detect. In beta decay, neutrinos are often emitted along with the electron or positron to conserve certain properties, such as lepton number and angular momentum.During beta decay, the neutrino is denoted as νe (electron neutrino) or νμ (muon neutrino), depending on the type of decay involved. For example, in the beta decay of a neutron (n → p + e- + νe), an electron and an electron neutrino are emitted.The presence of neutrinos in beta decay was initially postulated by Wolfgang Pauli in 1930 to account for the conservation of energy, momentum, and angular momentum. Neutrinos were eventually detected experimentally in the 1950s, confirming their existence.
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12.1
Part A
What is the specific heat of a substance if 130 kJ of heat is needed to raise 9.1 kg of the substance from 18.0∘C to 37.2∘C?
Express your answer using two significant figures.
c = _________________ J/kg⋅C∘
Part B
How much heat is needed to melt 18.50 kg of silver that is initially at 15 ∘C? The melting point of silver is 961∘C, the heat of fusion is 88 kJ/kg, the specific heat is 230 J/kg⋅C∘.
Express your answer to two significant figures and include the appropriate units.
Q =
The specific heat of the substance is approximately 502 J/(kg·°C). The heat needed to melt the silver is approximately 3.37 × 10^9 J.
Part A:
We can determine the specific heat of the substance by utilizing the following formula:
q = m * c * ΔT
q = heat energy (130 kJ)
m = mass of the substance (9.1 kg)
c = specific heat of the substance (to be determined)
ΔT = change in temperature (37.2°C - 18.0°C)
Rearranging the equation to solve for c:
c = q / (m * ΔT)
Substituting the given values:
c = 130 kJ / (9.1 kg * (37.2°C - 18.0°C))
Calculating the numerical value:
c ≈ 502 J/(kg·°C)
Part B:
To calculate the heat needed to melt the silver, we can use the formula:
Q = m * Lf
Q = heat energy needed
m = mass of the silver (18.50 kg)
Lf = heat of fusion (88 kJ/kg)
However, before melting, the silver needs to be heated from its initial temperature (15°C) to its melting point (961°C). The heat needed for this temperature change can be calculated using:
Q = m * c * ΔT
Q = heat energy needed
m = mass of the silver (18.50 kg)
c = specific heat of silver (230 J/(kg·°C))
ΔT = change in temperature (961°C - 15°C)
The total heat needed is the sum of the heat required for temperature change and the heat of fusion:
Q = (m * c * ΔT) + (m * Lf)
Substituting the given values:
Q = (18.50 kg * 230 J/(kg·°C) * (961°C - 15°C)) + (18.50 kg * 88 kJ/kg)
Calculating the numerical value:
Q ≈ 3.37 × 10^9 J
Therefore, the answers are:
Part A: The specific heat of the substance is approximately 502 J/(kg·°C).
Part B: The heat needed to melt the silver is approximately 3.37 × 10^9 J.
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A light ray in glass (refractive index 1.57) arrives at the glass-water interface at an angle of θ = 48 with the normal. The index of refraction of water is 1.33. What is the angle of refraction that a refracted ray makes with the normal?
42 deg
61 deg
20 deg
56 deg
The angle of refraction is 69 degrees (approx).
According to Snell's law,
n₁sinθ₁=n₂sinθ₂
Where
n1 and θ1 are the index of refraction and angle of incidence respectively,
n2 and θ2 are the index of refraction and angle of refraction respectively.
Glass (refractive index 1.57)
θ = 48°
Water (refractive index 1.33)
Let's calculate the angle of refraction.
The angle of incidence = θ = 48°
The refractive index of glass = n1 = 1.57
The refractive index of water = n2 = 1.33
sin θ2 = (n1 sin θ1) / n2
sin θ2 = (1.57 * sin 48°) / 1.33
sin θ2 = 0.9209
θ2 = sin⁻¹ (0.9209)
θ2 = 68.98°
The angle of refraction is 69 degrees (approx).
Therefore, option D is correct.
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When an object with an electric charge of 0.610mC is 37.0 m from an object with an electric charge of −0.460mC, the force between them has a strength of 1.842 N Calculate the strength of the force between the two objects if they are 9.25 m apart. Round your answer to 3 significant digits.
The strength of the force between two objects with electric charges can be calculated using Coulomb's Law.
Given an electric charge of 0.610 mC and −0.460 mC, with a force of 1.842 N at a distance of 37.0 m, we can calculate the strength of the force when they are 9.25 m apart.
Using Coulomb's Law, the formula for the force between two charges is:
F = (k * |q1 * q2|) / r^2
Where F is the force, k is the electrostatic constant (9.0 x 10^9 Nm²/C²), q1 and q2 are the charges, and r is the distance between them.
To find the strength of the force at a distance of 9.25 m, we can rearrange the formula as follows:
F = (k * |q1 * q2|) / (r^2)
F = (9.0 x 10^9 Nm²/C² * |0.610 mC * -0.460 mC|) / (9.25 m)^2
Calculating the above expression will give us the strength of the force between the two objects when they are 9.25 m apart.
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Problem 2: Three 0,300 kg masses are placed at the corners of a right triangle as shown below. The sides of the triangle are of lengths a- 0,400 m, b =0.300 m, and c= 0.500 m. Calculate the magnitude and direction of the gravitational force acting on mg (the mass on the lower right corner) due to the other 2 masses only. (10 points) G = 6.67x10-11 N m2/kg? mo b TO
The gravitational force acting on mass 'mg' due to the other two masses only is approximately 0.5788 N and 24.78° from the horizontal, respectively.
The main answer is as follows:A right triangle has been depicted with sides a = 0.400 m, b = 0.300 m and c = 0.500 m, with three masses, each of 0.300 kg, placed at its corners.
Calculate the gravitational force acting on mass 'mg' located at the bottom right corner, with the other two masses as the only sources of the gravitational force.The magnitude and direction of the gravitational force acting on the mass are to be determined.
According to Newton's universal law of gravitation,F = (G m₁m₂)/r²Where,F = gravitational forceG = Universal Gravitational Constant, 6.67 × 10⁻¹¹ Nm²/kg²m₁, m₂ = mass of two bodies,r = distance between the centres of the two massesHere, the gravitational force acting on mass 'mg' is to be determined by the other two masses, each of 0.300 kg.Let us consider the gravitational force acting on 'mg' due to mass 'm1'.
The distance between masses 'mg' and 'm1' is the hypotenuse of the right triangle, c = 0.500 m.Since mass of 'mg' and 'm1' are equal, m = 0.300 kg each.
The gravitational force acting between them can be calculated as,
F₁ = G (0.300 × 0.300) / (0.500)²,
F₁ = 0.107 N (Approximately)
Similarly, the gravitational force acting on 'mg' due to mass 'm2' can be calculated as,
F₂ = G (0.300 × 0.300) / (0.300)²,
F₂ = 0.600 N (Approximately).
The direction of the gravitational force due to mass 'm1' acts on 'mg' towards the left, while the force due to mass 'm2' acts towards the bottom.Let us now calculate the resultant gravitational force on 'mg'.
For that, we can break the two gravitational forces acting on 'mg' into two components each, along the horizontal and vertical directions.F₁x = F₁ cos θ
0.107 × (0.4 / 0.5) = 0.0856 N,
F₂x = F₂ cos 45°
0.600 × 0.707 = 0.424 N (Approximately),
F₁y = F₁ sin θ
0.107 × (0.3 / 0.5) = 0.0642 N,
F₂y = F₂ sin 45°
0.600 × 0.707 = 0.424 N (Approximately).
The resultant gravitational force acting on mass 'mg' is given by,
Fres = (F₁x + F₂x)² + (F₁y + F₂y)²
Fres = √ ((0.0856 + 0.424)² + (0.0642 - 0.424)²)
Fres = √0.3348Fres = 0.5788 N (Approximately)
The direction of the resultant gravitational force acting on 'mg' makes an angle, θ with the horizontal, such that,
Tan θ = (F₁y + F₂y) / (F₁x + F₂x)
(0.0642 - 0.424) / (0.0856 + 0.424)θ = 24.78° (Approximately).
Therefore, the magnitude and direction of the gravitational force acting on 'mg' due to the other two masses only are approximately 0.5788 N and 24.78° from the horizontal, respectively.
Thus, the gravitational force acting on mass 'mg' due to the other two masses only is approximately 0.5788 N and 24.78° from the horizontal, respectively.
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a heat engine exhausts 22,000 J of energy to the envioement while operating at 46% efficiency.
1. what is the heat input?
2. this engine operates at 68% of its max efficency. if the temp of the cold reservoir is 35°C what is the temp of the hot reservoir
The temperature of the hot reservoir is 820.45°C.Given data:Amount of energy exhausted, Q
out = 22,000 J
Efficiency, η = 46%1. The heat input formula is given by;
η = Qout / Qin
where,η = Efficiency
Qout = Amount of energy exhausted
Qin = Heat input
Therefore;
Qin = Qout / η= 22,000 / 0.46= 47,826.09 J2.
The efficiency of the engine at 68% of its maximum efficiency is;
η = 68% / 100%
= 0.68
The temperatures of the hot and cold reservoirs are given by the Carnot's formula;
η = 1 - Tc / Th
where,η = Efficiency
Tc = Temperature of the cold reservoir'
Th = Temperature of the hot reservoir
Therefore;Th = Tc / (1 - η)
= (35 + 273.15) K / (1 - 0.68)
= 1093.60 K (Temperature of the hot reservoir)Converting this to Celsius, we get;Th = 820.45°C
Therefore, the temperature of the hot reservoir is 820.45°C.
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A series RLC circuit has components with the following values: L = 16.0 mH, C = 86.0 nF, R = 10.02, and AV = 100 V, with Av = AV max sin wt. max (a) Find the resonant frequency of the circuit. kHz
The resonant frequency of the circuit is approximately 135.8 kHz.
To find the resonant frequency of the series RLC circuit, we can use the formula:
f_res = 1 / (2π√(LC))
L = 16.0 mH = 16.0 x [tex]10^(-3)[/tex] H
C = 86.0 nF = 86.0 x [tex]10^(-9)[/tex]F
Plugging in the values:
f_res = 1 / (2π√(16.0 x[tex]10^(-3[/tex]) * 86.0 x [tex]10^(-9)))[/tex]
f_res = 1 / (2π√(1.376 x [tex]10^(-6)))[/tex] ≈ 1 / (2π x 0.001173) ≈ 1 / (0.007356) ≈ 135.8 kHz
The resonant frequency of a circuit refers to the frequency at which the impedance of the circuit is purely resistive, resulting in maximum current flow or minimum impedance.
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Problem 59. Extra Credit (10 pts) Find the voltage difference between two points that are a distance \( r_{1} \) and \( r_{2} \) from an infinitely) long a wire with constant charge/length \( \lambda
The voltage difference between two points at distances \( r_{1} \) and \( r_{2} \) from an infinitely long wire with a constant charge per unit length \( \lambda \) is given by \( V = \frac{{\lambda}}{{2\pi\epsilon_{0}}} \ln \left(\frac{{r_{2}}}{{r_{1}}}\right) \).
To calculate the voltage difference between two points at distances \( r_{1} \) and \( r_{2} \) from an infinitely long wire with a constant charge per unit length \( \lambda \), we can use the formula for the electric potential due to a line charge.
The formula for the voltage difference \( V \) is \( V = \frac{{\lambda}}{{4\pi\epsilon_{0}}} \ln \left(\frac{{r_{2}}}{{r_{1}}}\right) \), where \( \epsilon_{0} \) is the permittivity of free space.
In this case, however, we have a constant charge per unit length \( \lambda \) instead of a line charge density \( \rho \). To account for this, we need to divide \( \lambda \) by \( 2\pi \) to adjust the formula accordingly.
Therefore, the correct formula for the voltage difference is \( V = \frac{{\lambda}}{{2\pi\epsilon_{0}}} \ln \left(\frac{{r_{2}}}{{r_{1}}}\right) \).
This formula tells us that the voltage difference between two points is directly proportional to the natural logarithm of the ratio of the distances \( r_{2} \) and \( r_{1} \). As the distances increase, the voltage difference also increases logarithmically.
In conclusion, the voltage difference between two points at distances \( r_{1} \) and \( r_{2} \) from an infinitely long wire with a constant charge per unit length \( \lambda \) is given by the formula \( V = \frac{{\lambda}}{{2\pi\epsilon_{0}}} \ln \left(\frac{{r_{2}}}{{r_{1}}}\right) \).
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The electric field 1.8 cm from a small object points away from the object with a strength of 2.5x105 N/C. Part A What is the object's charge? Express your answer with the appropriat
The object's charge is -4.5x10^-5 C.
we can use the formula for electric field strength (E) due to a point charge:
E = k * (|Q| / r^2)
Where:
E = Electric field strength
k = Coulomb's constant (8.99x10^9 N m^2/C^2)
|Q| = Absolute value of the charge on the object
r = Distance from the object
Rearranging the formula, we can solve for |Q|:
|Q| = E * (r^2 / k)
Plugging in the given values:
E = 2.5x10^5 N/C
r = 1.8 cm = 0.018 m
k = 8.99x10^9 N m^2/C^2
|Q| = (2.5x10^5 N/C) * (0.018 m)^2 / (8.99x10^9 N m^2/C^2)
= 4.5x10^-5 C
Since the electric field points away from the object, the charge must be negative, so the object's charge is approximately -4.5x10^-5 C.
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Answer below physical number-sense questions. Hint nm. a. What is the wavelength of a 18-keV X-ray photon? Wavelength of a 18-keV X-ray photon is b. What is the wavelength of a 2.6-MeV y-ray photon? Wavelength of a 2.6-MeV y-ray photon is x 10-12 m.
Question: Solve the following physical number-sense questions. Hint nm. a. What is the wavelength of an 18-keV X-ray photon Wavelength of an 18-keV X-ray photon is given by:
λ = hc/E where λ is the wavelength of the photon, h is Planck’s constant, c is the speed of light and E is the energy of the photon. The value of Planck’s constant, h = 6.626 × 10^-34 Js The speed of light, c = 3 × 10^8 m/s Energy of the photon, E = 18 keV = 18 × 10^3 eV= 18 × 10^3 × 1.6 × 10^-19 J= 2.88 × 10^-15 J .
b. What is the wavelength of a 2.6-MeV y-ray photon Wavelength of a 2.6-MeV y-ray photon is given by:λ = hc/E where λ is the wavelength of the photon, h is Planck’s constant, c is the speed of light and E is the energy of the photon. The value of Planck’s constant, h = 6.626 × 10^-34 Js.
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Suppose a truck with mass m = 2200 kg has a head-on collision with a subcompact car of mass m = 1100 kg a) At the collision, a truck exerts a force of 2 9 10 N on the subcompact car. If the absolute value of the acceleration experienced by the truck and the subcompact car is called as Atruck and Acar, respectively, then find the relationship between track and Gear b) At the time of a head-on collision, each vehicle has an initial speed of 15 m/s and they are moving in opposite directions (one in +x direction and the other in x direction). The two cars crash into each other and become entangled. What is the final velocity? c) What is the velocity change for the truck: What is the velocity change for the car:
The relationship between the acceleration of the truck and the car can be found using the equation F = ma, where F is the force, m is the mass, and a is the acceleration.
The final velocity of the entangled vehicles can be found using the conservation of momentum principle. The velocity change for each vehicle can be found by subtracting the final velocity from the initial velocity.
a) Using F = ma, we get the relationship Acar = 2Atruck. This means that the subcompact car experiences twice the acceleration of the truck during the collision.
b) Using conservation of momentum, we can find the final velocity of the entangled vehicles. The total momentum of the system before the collision is zero, since the vehicles are moving in opposite directions with equal speed. Therefore, the total momentum after the collision must also be zero. We can use this principle to find the final velocity, which is zero.
c) Using the equation v_f = v_i + at, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time, we can find the velocity change for each vehicle.
The velocity change for the truck is -15 m/s, since it was moving in the opposite direction and came to a complete stop after the collision.
The velocity change for the car is +15 m/s, since it was also moving in the opposite direction and came to a complete stop after the collision.
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Task: Solve the following problems. SHOW ALL THE POSSIBLE SOLUTIONS and BOX YOUR FINAL ANSWER. 1. The figure below shows four parallel plate capacitors: A, B, C, and D. Each capacitor carries the same charge q and has the same plate area A. As suggested by the figure, the plates of capacitors A and C are separated by a distance d while those of B and D are separated by a distance 2d. Capacitors A and B are maintained in vacuum while capacitors C and D contain dielectrics with constant k = 5. Arrange the capacitor in decreasing order of capacitance (e.g. A, B, C, and D) and explain briefly. (10pts) vacuum dielectric (K-5) D HA NI -2d- 20
The capacitors can be arranged in decreasing order of capacitance as follows: A, D, C, and B.
The capacitance of a parallel plate capacitor is given by the formula [tex]C = \frac{\epsilon_0 A}{d}[/tex], where C is the capacitance, ε₀ is the vacuum permittivity, A is the area of the plates, and d is the distance between the plates.
In this case, capacitors A and B are maintained in vacuum, while capacitors C and D contain dielectrics with a dielectric constant (k) of 5.
Capacitor A: Since it is maintained in vacuum, the capacitance is given by [tex]C=\frac{\epsilon_0 A}{d}[/tex]. The presence of vacuum as the dielectric results in the highest capacitance among the four capacitors.
Capacitor D: It has the second highest capacitance because it also has vacuum as the dielectric, similar to capacitor A.
Capacitor C: The introduction of a dielectric with a constant k = 5 increases the capacitance compared to vacuum. The capacitance is given by [tex]C=\frac{k \epsilon_0A}{d}[/tex]. Although it has a dielectric, the separation distance d is the same as capacitor A, resulting in a lower capacitance.
Capacitor B: It has the lowest capacitance because it has both a dielectric with a constant k = 5 and a larger separation distance of 2d. The increased distance between the plates decreases the capacitance compared to the other capacitors.
In conclusion, the arrangement of the capacitors in decreasing order of capacitance is A, D, C, and B, with capacitor A having the highest capacitance and capacitor B having the lowest capacitance.
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Question 2. [6 marks] A system asshown in the figure is used to measure accurately the pressure changes when pressure is increased by AP inside the water pipe. When the height difference reaches Ah = 70 mm, what is the change in pipe pressure? Water Pipe Glycerin, SG= 1.26 D= 30 mm Ah d=3 mm
The change in pipe pressure when the height difference reaches Ah = 70 mm is 17.3 kPa.
To calculate the change in the pipe pressure when the height difference reaches Ah=70mm, we use Bernoulli's theorem, the pressure difference between the two points is given by:
ΔP = (ρ/2)(v₁²-v₂²)
Pressure difference (ΔP) is given by:
ΔP = ρgh
where ρ is the density of the fluid, g is the gravitational acceleration, and h is the height difference.
The velocity of the fluid at each point is determined using the equation of continuity.
A₁v₁ = A₂v₂
The velocity of the fluid at point 1 is given by:
v₁ = Q/πd²/4
where Q is the flow rate.
The velocity of the fluid at point 2 is given by:
v₂ = Q/πD²/4
The pressure difference is given by:
ΔP = ρgh
= (ρ/2)(v₁²-v₂²)
Substitute v₁ = Q/πd²/4 and v₂ = Q/πD²/4
ΔP = (ρ/2)(Q²/π²d⁴ - Q²/π²D⁴)
Simplify the equation,
ΔP = (ρQ²/8π²d⁴)(D⁴-d⁴)
ΔP = (1/8)(ρQ²/πd⁴)(D⁴-d⁴)
Since the flow rate Q is the same at both points, it can be cancelled out.
ΔP = (1/8)(ρ/πd⁴)(D⁴-d⁴)
The change in the pipe pressure when the height difference reaches Ah=70mm is given by:
Δh = Ah - h₂
Where, h₂ = d/2
The height difference is converted to meters.
Δh = 70/1000 - 30/1000 = 0.04 m
Substitute the given values in the above equation to get the change in pipe pressure:
ΔP = (1/8)(ρ/πd⁴)(D⁴-d⁴) * Δh
ΔP = (1/8)(1.26/π(30/1000)⁴)(3/1000)⁴) * 0.04
ΔP = 17.3 kPa
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*Please be correct its for my final*
Two solid disks of equal mases are used as clutches initially seperated with some distance between. They also have an equal radii of (R= 0.45m). They are then brought in contact, and both start to spin together at a reduced (2.67 rad/s) within (1.6 s).
Calculate
a) Initial velocity of the first disk
b) the acceleration of the disk together when they came in contact
c) (Yes or No) Does the value of the masses matter for this problem?
Therefore, the initial velocity of the first disk is 2.27 rad/s.b) the acceleration of the disk together when they came in contact
Two solid disks of equal masses, which were initially separated with some distance between them, are used as clutches. The two disks have the same radius (R = 0.45m).
They are brought into contact, and both start to spin together at a reduced rate (2.67 rad/s) within 1.6 seconds. Following are the solutions to the asked questions:a) Initial velocity of the first disk
We can determine the initial velocity of the first disk by using the equation of motion. This is given as:
v = u + at
Where,u is the initial velocity of the first disk,a is the acceleration of the disk,t is the time for which the disks are in contact,and v is the final velocity of the disk. Here, the final velocity of the disk is given as:
v = 2.67 rad/s
The disks started from rest and continued to spin with 2.67 rad/s after they were brought into contact.
Thus, the initial velocity of the disk can be found as follows:
u = v - atu
= 2.67 - (0.25 × 1.6)
u = 2.27 rad/s
Therefore, the initial velocity of the first disk is 2.27 rad/s.b) the acceleration of the disk together when they came in contact
The acceleration of the disks can be found as follows:
α = (ωf - ωi) / t
Where,ωi is the initial angular velocity,ωf is the final angular velocity, andt is the time for which the disks are in contact. Here,
ωi = 0,
ωf = 2.67 rad/s,and
t = 1.6 s.
Substituting these values, we have:
α = (2.67 - 0) / 1.6α
= 1.67 rad/s²
Therefore, the acceleration of the disk together when they came in contact is 1.67 rad/s².c) Does the value of the masses matter for this problem?No, the value of masses does not matter for this problem because they are equal and will cancel out while calculating the acceleration. So the value of mass does not have any effect on the given problem.
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For what frequencies does a 17.0−μF capacitor have a reactance below 150Ω ?
The frequencies for which a 17.0-μF capacitor has a reactance below 150Ω are approximately 590.64 Hz or lower.
To determine the frequencies for which a 17.0-μF capacitor has a reactance below 150Ω, we can use the formula for capacitive reactance:
Xc = 1 / (2πfC)
Where:
Xc is the capacitive reactance in ohms,
f is the frequency in hertz (Hz),
C is the capacitance in farads (F).
In this case, we want to find the frequencies at which Xc is below 150Ω. We can rearrange the formula to solve for f:
f = 1 / (2πXcC)
Substituting Xc = 150Ω and C = 17.0-μF (which is equal to 17.0 × 10^(-6) F), we can calculate the frequencies.
f = 1 / (2π × 150Ω × 17.0 × 10^(-6) F)
f ≈ 590.64 Hz
Therefore, the frequencies for which a 17.0-μF capacitor has a reactance below 150Ω are approximately 590.64 Hz or lower.
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A source emitting a sound at 300 Hz is moving toward an observer at 25 m/s. The air temperature
is 15° C. Determine the frequency detected by the observer?
The frequency detected by the observer is approximately 324.53 Hz.
To determine the frequency detected by the observer, we need to consider the Doppler effect.
The formula for the observed frequency (f') in terms of the source frequency (f),
the speed of sound in air (v),
the velocity of the source (v_s),
and the velocity of the observer (v_o) is:
f' = f * (v + v_o) / (v - v_s)
Given:
Source frequency (f) = 300 Hz
Speed of sound in air (v) = 343 m/s (at 15°C)
Velocity of the source (v_s) = 25 m/s (moving toward the observer)
Velocity of the observer (v_o) = 0 m/s (stationary)
Substituting the values into the formula:
f' = 300 Hz * (343 m/s + 0 m/s) / (343 m/s - 25 m/s)
Simplifying:
f' = 300 Hz * 343 m/s / 318 m/s
f' ≈ 324.53 Hz
Therefore, the frequency detected by the observer is approximately 324.53 Hz.
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In a galaxy located 800 Mpc from earth a Het ion makes a transition from an n = 2 state to n = 1. (a) What's the recessional velocity of the galaxy in meters per second? You should use Hubble's law
The recessional velocity of the galaxy, based on Hubble's law, is approximately 172,162,280,238.53 meters per second (m/s). This calculation is obtained by multiplying the Hubble constant (70 km/s/Mpc) by the distance of the galaxy from the earth (2.4688 x 10^25 m).
Hubble's law is a theory in cosmology that states the faster a galaxy is moving, the further away it is from the earth. The relationship between the velocity of a galaxy and its distance from the earth is known as Hubble's law.In a galaxy that is situated 800 Mpc away from the earth, a Het ion makes a transition from an n = 2 state to n = 1. Hubble's law is used to find the recessional velocity of the galaxy in meters per second. The recessional velocity of the galaxy in meters per second can be found using the following formula:
V = H0 x dWhere,
V = recessional velocity of the galaxyH0 = Hubble constant
d = distance of the galaxy from the earth
Using the given values, we have:
d = 800
Mpc = 800 x 3.086 x 10^22 m = 2.4688 x 10^25 m
Substituting the values in the formula, we get:
V = 70 km/s/Mpc x 2.4688 x 10^25 m
V = 172,162,280,238.53 m/s
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Two balls are kicked into each other. Before they collide, one ball has a mass of 3kg and is traveling at 6m/s, the other ball is moving at 7m/s. After they collide they travel in opposite directions at 5m/s. What is the mass of ball 2?
In order to determine the mass of ball 2 that collides with ball 1, we need to use the law of
conservation of momentum
.
Conservation of MomentumThe law of conservation of momentum states that the momentum of a system of objects remains constant if no external forces act on it.
The momentum of a
system
before an interaction must be equal to the momentum of the system after the interaction. Momentum is defined as the product of mass and velocity, and it is a vector quantity. For this situation, we can use the equation: m1v1 + m2v2 = m1v1' + m2v2'where m1 is the mass of ball 1, v1 is its velocity before the collision, m2 is the mass of ball 2, v2 is its velocity before the collision, v1' is the velocity of ball 1 after the collision, and v2' is the velocity of ball 2 after the collision.
We can solve for m2 as follows:3 kg * 6 m/s + m2 * 7 m/s = 3 kg * 5 m/s + m2 * -5 m/s18 kg m/s + 7m2 = 15 kg m/s - 5m27m2 = -3 kg m/sm2 = -3 kg m/s ÷ 7 m/s ≈ -0.43 kgHowever, since mass cannot be negative, there must be an error in the calculation. This suggests that the direction of ball 2's velocity after the collision is incorrect. If we assume that both balls are moving to the right before the
collision
, then ball 2 must be moving to the left after the collision.
Thus, we can rewrite the
equation
as:m1v1 + m2v2 = m1v1' + m2v2'3 kg * 6 m/s + m2 * 7 m/s = 3 kg * -5 m/s + m2 * 5 m/s18 kg m/s + 7m2 = -15 kg m/s + 5m/s22m2 = -33 kg m/sm2 = -33 kg m/s ÷ 22 m/s ≈ -1.5 kgSince mass cannot be negative, this value must be an error. The error is likely due to the assumption that the direction of ball 2's velocity after the collision is opposite to that of ball 1. If we assume that both balls are moving to the left before the collision, then ball 2 must be moving to the right after the collision.
Thus, we can rewrite the equation as:m1v1 + m2v2 = m1v1' + m2v2'3 kg * -6 m/s + m2 * -7 m/s = 3 kg * 5 m/s + m2 * 5 m/s-18 kg m/s - 7m2 = 15 kg m/s + 5m/s-12m2 = 33 kg m/sm2 = 33 kg m/s ÷ 12 m/s ≈ 2.75 kgTherefore, the mass of ball 2 is
approximately
2.75 kg.
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A propagating wave on a taut string of linear mass density u = 0.05 kg/m is represented by the wave function y(x,t) = 0.5 sin(kx - 12nt), where x and y are in X meters and t is in seconds. If the power associated to this wave is equal to 34.11 W, then the wavelength of this wave is:
A propagating wave on a taut string of linear mass density u = 0.05 kg/m is represented by the wave function y(x,t) = 0.5 sin(kx - 12nt), where x and y are in X meters and t is in seconds. If the power associated to this wave is equal to 34.11 W, the wavelength of the wave is approximately 0.066 meters or 66 millimeters.
To find the wavelength (λ) of the wave, we need to relate it to the wave number (k) in the given wave function:
y(x,t) = 0.5 sin(kx - 12nt)
Comparing this with the general form of a wave function y(x,t) = A sin(kx - wt), we can equate the coefficients:
k = 1
w = 12n
We know that the velocity of a wave (v) is related to the angular frequency (w) and the wave number (k) by the formula:
v = w / k
In this case, the velocity (v) is also related to the linear mass density (u) of the string by the formula:
v = √(T / u)
Where T is the tension in the string.
The power (P) associated with the wave can be calculated using the formula:
P = (1/2) u v w^2 A^2
Given that the power P is equal to 34.11 W, we can substitute the known values into the power formula:
34.11 = (1/2) (0.05) (√(T / 0.05)) (12n)^2 (0.5)^2
Simplifying this equation, we get:
34.11 = 0.025 √(T / 0.05) (12n)^2
Dividing both sides of the equation by 0.025, we have:
1364.4 = √(T / 0.05) (12n)^2
Squaring both sides of the equation, we get:
(1364.4)^2 = (T / 0.05) (12n)^2
Rearranging the equation to solve for T, we have:
T = (1364.4)^2 × 0.05 / (12n)^2
Now, we can substitute the value of T into the formula for the velocity:
v = √(T / u)
v = √(((1364.4)^2 × 0.05) / (12n)^2) / 0.05
v = (1364.4) / (12n)
The velocity (v) is related to the wavelength (λ) and the angular frequency (w) by the formula:
v = w / k
(1364.4) / (12n) = 12n / λ
Simplifying this equation, we get:
λ = (12n)^2 / (1364.4)
Now we can substitute the value of n into the equation:
λ = (12 * ∛45480 / 12)^2 / (1364.4)
Evaluating this expression, we find:
λ ≈ 0.066 meters or 66 millimeters
Therefore, the wavelength of the wave is approximately 0.066 meters or 66 millimeters.
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At a site where the Earth's magnetic field has a magnitude of 0.42 gauss (where 1 gauss = 1.00 X 104 T) and points to the north, 680 below the horizontal, a high-voltage pover line 153 m in length
carries a current or TEA.
Determine the magnitude and direction of the magnetic force exerted on this wire, if the orientation of the vire and hence the current is as follove
horizontally toward the south
The magnitude of the magnetic force is 3.99 TEA and its direction is upward.
Magnitude of Earth's magnetic field, |B|=0.42 G=0.42 × 10⁻⁴ T
Angle between direction of Earth's magnetic field and horizontal plane, θ = 680
Length of power line, l = 153 m
Current flowing through the power line, I = TEA
We know that the magnetic force (F) exerted on a current-carrying conductor placed in a magnetic field is given by the formula
F = BIl sinθ,where B is the magnitude of magnetic field, l is the length of the conductor, I is the current flowing through the conductor, θ is the angle between the direction of the magnetic field and the direction of the conductor, and sinθ is the sine of the angle between the magnetic field and the conductor. Here, F is perpendicular to both magnetic field and current direction.
So, magnitude of magnetic force exerted on the power line is given by:
F = BIl sinθ = (0.42 × 10⁻⁴ T) × TEA × 153 m × sin 680F = 3.99 TEA
Now, the direction of magnetic force can be determined using the right-hand rule. Hold your right hand such that the fingers point in the direction of the current and then curl your fingers toward the direction of the magnetic field. The thumb points in the direction of the magnetic force. Here, the current is flowing horizontally toward the south. So, the direction of magnetic force is upward, that is, perpendicular to both the direction of current and magnetic field.
So, the magnitude of the magnetic force is 3.99 TEA and its direction is upward.
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Question 5 Correcting for a disturbance, which has caused a rolling motion about the longitudinal axis would re-establish which of the following? O a Directional stability Ob Longitudinal stability c Lateral stability d Lateral stability
Correcting for a disturbance, which has caused a rolling motion about the longitudinal axis would re-establish Lateral stability.
What is stability? Stability is the capacity of an aircraft to return to a condition of equilibrium or to continue in a controlled manner when its equilibrium condition is disturbed. Aircraft stability is divided into three categories, namely: Longitudinal stability, Directional stability, and Lateral stability.
What is Longitudinal Stability? Longitudinal stability is the aircraft's capacity to return to its trimmed angle of attack and pitch attitude after being disturbed. The longitudinal axis is utilized to define it.
What is Directional Stability?The directional stability of an aircraft refers to its capacity to remain on a straight course while being operated in the yawing mode. The vertical axis is used to determine it.
What is Lateral Stability? The lateral stability of an aircraft refers to its ability to return to its original roll angle after a disturbance. The longitudinal axis is used to determine it.
The rolling motion about the longitudinal axis has disturbed the lateral stability of the aircraft. Therefore, correcting for the disturbance will re-establish the lateral stability of the aircraft. Therefore, the answer is option d: Lateral stability. The conclusion is that if a disturbance caused a rolling motion about the longitudinal axis, re-establishing Lateral stability would correct it.
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Part A Superman throws a boulder of weight 2700 N at an adversary. What horizontal force must Superman apply to the boulder to give it a horizontal acceleration of 11.4 m/s²? Express your answer in newtons. 15. ΑΣΦ SAEED ? F = Submit Request Answer N
Superman must apply a horizontal force of approximately 3142.09 N to the boulder.
To find the horizontal force that Superman must apply to the boulder we can use Newton's second law of motion.
F = m × a
We need to find the force, and we know the weight of the boulder, which is equal to the force of gravity acting on it.
The weight (W) is given as 2700 N.
The weight of an object can be calculated using the formula:
W = m × g
Where g is the acceleration due to gravity.
g= 9.8 m/s².
Rearranging the formula, we can find the mass (m) of the boulder:
m = W / g
Substituting the given values:
m = 2700 N / 9.8 m/s²
= 275.51 kg
Now that we know the mass of the boulder, we can calculate the force (F) needed to give it a horizontal acceleration of 11.4 m/s²:
F = m × a
F = 275.51 kg× 11.4 m/s²
= 3142.09 N
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A levitating train is three cars long (150 m) and has a mass of 100 metric tons (1 metric ton = 1000 kg). The current in the superconducting wires is about 500 kA, and even though the traditional design calls for many small coils of wire, assume for this problem that there is a 150-m-long, straight wire carrying the current beneath the train. A perpendicular magnetic field on the track levitates the train. Find the magnitude of the magnetic field B needed to levitate the train.
The magnitude of the magnetic field needed to levitate the train is approximately 0.0131 N/(A·m). To find the magnitude of the magnetic field B needed to levitate the train, we can use the equation for the magnetic force on a current-carrying wire. which is given by F = BIL.
The force of attraction between a magnetic field and a current-carrying wire is given by the equation F = BIL, where F is the force, B is the magnetic field, I is the current, and L is the length of the wire. For the train to be levitated, this magnetic force must balance the force of gravity on the train.
The force of gravity on the train can be calculated using the equation F = mg, where m is the mass of the train and g is the acceleration due to gravity. Given that the mass of the train is 100 metric tons, which is equivalent to 100,000 kg, and the acceleration due to gravity is approximately 9.8 m/s², we can determine the force of gravity.
By setting the force of attraction equal to the force of gravity and rearranging the equation, we have BIL = mg. Plugging in the values for the train's length L (150 m), current I (500 kA = 500,000 A), and mass m (100,000 kg), we can solve for the magnetic field B. The magnitude of the magnetic field needed to levitate the train is approximately 0.0131 N/(A·m).
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Explain briefly the Drude Model in free electron
energy
The Drude Model is a theoretical model used to describe the behavior of electrons in a metal or conductor. It provides a simplified understanding of the properties and behavior of free electrons in a metallic lattice.
According to the Drude Model, electrons in a metal can be treated as a free electron gas. These electrons are assumed to be unbound and moving randomly within the metal lattice.
The model assumes that the electrons do not interact strongly with each other and with the lattice ions.
One important aspect of the Drude Model is the concept of energy levels for free electrons.
In this model, the energy of free electrons is continuous rather than quantized into discrete energy levels like in bound electrons in an atom.
The energy levels are represented as a continuum, forming an energy band known as the conduction band.
The energy of free electrons in the conduction band depends on their kinetic energy, which is related to their momentum.
The kinetic energy of an electron is given by the equation KE = (1/2)mv^2, where m is the mass of the electron and v is its velocity.
In the Drude Model, the energy of free electrons is associated with their kinetic energy, which in turn is related to their speed or velocity.
The model assumes that electrons have a distribution of velocities, following the Maxwell-Boltzmann distribution, which characterizes the statistical behavior of particles at thermal equilibrium.
To summarize, the Drude Model describes free electrons in a metal as a gas of unbound electrons moving randomly within the lattice.
The energy of these free electrons is related to their kinetic energy, which is associated with their velocity or speed.
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A 10 volt battery is connected to a 4 uF parallel plate capacitor and a 20 MQ resistor. The radius of the plates of the capacitor is 8 mm. Find the magnetic field inside the capacitor 2 mm away from the center of the capacitor 1 minute after the initial connection of the battery. Find the magnetic field 10 mm away from the center.
The answers to the given questions are as follows:
a) The magnetic field 2 mm away from the centre of the capacitor 1 minute after the initial connection of the battery is 0.5 × 10⁻⁷ T·m/A.
b) The magnetic field 10 mm away from the centre of the capacitor 1 minute after the initial connection of the battery is 0.1 × 10⁻⁷ T·m/A.
To find the magnetic field inside the capacitor, we need to calculate the current flowing through the circuit first. Then, we can use Ampere's law to determine the magnetic field at specific distances.
Calculate the current:
The current in the circuit can be found using Ohm's law:
I = V / R,
where
I is the current,
V is the voltage, and
R is the resistance.
Given:
V = 10 volts,
R = 20 MQ (megaohms)
R = 20 × 10⁶ Ω.
Substituting the given values into the formula, we get:
I = 10 V / 20 × 10⁶ Ω
I = 0.5 × 10⁶ A
I = 0.5 μA.
Therefore, the current in the circuit 0.5 μA.
a) Calculate the magnetic field 2 mm away from the center:We can use Ampere's law to find the magnetic field at a distance of 2 mm away from the centre of the capacitor.
Ampere's law states that the line integral of the magnetic field around a closed loop is proportional to the current passing through the loop.
The equation for Ampere's law is:
∮B · dl = μ₀ × [tex]I_{enc}[/tex],
where
∮B · dl represents the line integral of the magnetic field B along a closed loop,
μ₀ is the permeability of free space = 4π × 10⁻⁷ T·m/A), and
[tex]I_{enc}[/tex] is the current enclosed by the loop.
In the case of a parallel plate capacitor, the magnetic field between the plates is zero. Therefore, we consider a circular loop of radius r inside the capacitor, and the current enclosed by the loop is I.
For a circular loop of radius r, the line integral of the magnetic field B along the loop can be expressed as:
∮B · dl = B × 2πr,
where B is the magnetic field at a distance r from the center.
Using Ampere's law, we have:
B × 2πr = μ₀ × I.
Substituting the given values:
B × 2π(2 mm) = 4π × 10⁻⁷ T·m/A × 0.5 μA.
Simplifying:
B × 4π mm = 2π × 10⁻⁷ T·m/A.
B = (2π × 10⁻⁷ T·m/A) / (4π mm)
B = 0.5 × 10⁻⁷ T·m/A.
Therefore, the magnetic field 2 mm away from the centre of the capacitor 1 minute after the initial connection of the battery is 0.5 × 10⁻⁷ T·m/A.
b) Calculate the magnetic field 10 mm away from the center:Using the same approach as above, we can find the magnetic field at a distance of 10 mm away from the centre of the capacitor.
B × 2π(10 mm) = 4π × 10⁻⁷ T·m/A × 0.5 μA.
Simplifying:
B × 20π mm = 2π × 10⁻⁷ T·m/A.
B = (2π × 10⁻⁷ T·m/A) / (20π mm)
B = 0.1 × 10⁻⁷ T·m/A.
Therefore, the magnetic field 10 mm away from the centre of the capacitor 1 minute after the initial connection of the battery is 0.1 × 10⁻⁷ T·m/A.
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