To calculate the strength of the magnetic field at point P in the given figure, we can use Ampere's Law. Ampere's Law states that the line integral of the magnetic field around a closed loop is equal to the product of the permeability of free space (μ₀) and the current enclosed by the loop.
In this case, the loop can be chosen as a circle centered at point P with a radius equal to r2. The current enclosed by the loop is I.
Using Ampere's Law, we have:
∮ B · dl = μ₀ * I_enclosed
Since the magnetic field is assumed to be constant along the circular path, we can simplify the equation to:
B * 2πr2 = μ₀ * I
Solving for B, we get:
B = (μ₀ * I) / (2πr2)
Plugging in the given values:
B = (4π × 10^-7 T·m/A) * (5.6 A) / (2π × 0.028 m)
B ≈ 0.04 T
Therefore, the strength of the magnetic field at point P is approximately 0.04 Tesla.
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25 A plank AB 3.0 m long weighing 20 kg and with its centre of gravity 2.0 m from the end A carries a load of mass 10 kg at the end A. It rests on two supports at C and D as shown in fig. 4.48. R₁ A A C 50 cm 10 kg Fig. 4.49 (i) 2.0 m R₂ D 50 cm B 10 Fi 28 Compute the values of the reaction 29 forces R₁ and R₂ at C and D.
(1) R1 = 294 N, R2 = 588 N.
(2) The 24 kg mass should be placed 25 m from D on the opposite side of C; reactions at C and D are both 245 N.
(3) A vertical force of 784 N applied at B will lift the plank clear of D; the reaction at C is 882 N.
To solve this problem, we need to apply the principles of equilibrium. Let's address each part of the problem step by step:
(1) To calculate the reaction forces R1 and R2 at supports C and D, we need to consider the rotational equilibrium and vertical equilibrium of the system. Since the plank is in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments. Taking moments about point C, we have:
Clockwise moments: (20 kg × 9.8 m/s² × 20 m) + (10 kg × 9.8 m/s² × 30 m)
Anticlockwise moments: R2 × 3.0 m
Setting the moments equal, we can solve for R2:
(20 kg × 9.8 m/s² × 20 m) + (10 kg × 9.8 m/s² × 30 m) = R2 × 3.0 m
Solving this equation, we find R2 = 588 N.
Now, to find R1, we can use vertical equilibrium:
R1 + R2 = 20 kg × 9.8 m/s² + 10 kg × 9.8 m/s²
Substituting the value of R2, we get R1 = 294 N.
Therefore, R1 = 294 N and R2 = 588 N.
(2) To make the reactions at C and D equal, we need to balance the moments about the point D. Let x be the distance from D to the 24 kg mass. The clockwise moments are (20 kg × 9.8 m/s² × 20 m) + (10 kg × 9.8 m/s² × 30 m), and the anticlockwise moments are 24 kg × 9.8 m/s² × x. Setting the moments equal, we can solve for x:
(20 kg × 9.8 m/s² × 20 m) + (10 kg × 9.8 m/s² × 30 m) = 24 kg × 9.8 m/s² × x
Solving this equation, we find x = 25 m. The mass of 24 kg should be placed 25 m from D on the opposite side of C.
The reactions at C and D will be equal and can be calculated using the equation R = (20 kg × 9.8 m/s² + 10 kg × 9.8 m/s²) / 2. Substituting the values, we get R = 245 N.
(3) Without the 24 kg mass, to lift the plank clear of D, we need to consider the rotational equilibrium about D. The clockwise moments will be (20 kg × 9.8 m/s² × 20 m) + (10 kg × 9.8 m/s² × 30 m), and the anticlockwise moments will be F × 3.0 m (where F is the vertical force applied at B). Setting the moments equal, we have:
(20 kg × 9.8 m/s² × 20 m) + (10 kg × 9.8 m/s² × 30 m) = F × 3.0 m
Solving this equation, we find F = 784 N.
The reaction at C can be calculated using vertical equilibrium: R1 + R2 = 20 kg × 9.8 m/s² + 10 kg × 9.8 m/s². Substituting the values, we get R1 + R2 = 294 N + 588 N = 882 N.
In summary, (1) R1 = 294 N and R2 = 588 N. (2) The 24 kg mass should be placed 25 m from D on the opposite side of C, and the reactions at C and D will be equal to 245 N. (3) Without the 24 kg mass, a vertical force of 784 N applied at B will lift the plank clear of D, and the reaction at C will be 882 N.
The question was incomplete. find the full content below:
A plank ab 3.0 long weighing20kg and with its centre gravity 20m from the end a carries a load of mass 10kg at the end a.It rests on two supports at c and d.Calculate:
(1)compute the values of the reaction forces R1 and R2 at c and d
(2)how far from d and on which side of it must a mass of 24kg be placed on the plank so as to make the reactions equal?what are their values?
(3)without this 24kg,what vertical force applied at b will just lift the plank clear of d?what is then the reaction of c?
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When the Venera 14 probe landed on Venus's surface, its barometer measured an air pressure of 9.5 MPa. The surface acceleration due to gravity was measured to be 8.87 m/s2. If Earth's atmosphere with a pressure of 101 kPa raises mercury 0.760 m where gravitational acceleration is 9.81 m/s2. To what height in m to two significant digits would Venus's atmosphere raise liquid mercury?
The height to which Venus's atmosphere would raise liquid mercury is determined based on the given air pressure and surface acceleration due to gravity. The calculation involves comparing the pressure in Venus's atmosphere to Earth's atmosphere and using the difference to determine the height of the mercury column.
To calculate the height to which Venus's atmosphere would raise liquid mercury, we can use the principle of hydrostatic pressure. The pressure difference between two points in a fluid column is directly proportional to the difference in height.Given that Earth's atmosphere raises mercury to a height of 0.760 m when the pressure is 101 kPa and the acceleration due to gravity is 9.81 m/s^2, we can set up a proportion to find the height in Venus's atmosphere.
The ratio of pressure to height is constant, so we can write:
(9.5 MPa / 101 kPa) = (8.87 m/s^2 / 9.81 m/s^2) * (h / 0.760 m)
Solving for h, we can find the height to which Venus's atmosphere would raise liquid mercury.
By rearranging the equation and substituting the given values, we can calculate the height to two significant digits.
Therefore, the height to which Venus's atmosphere would raise liquid mercury can be determined using the given air pressure and surface acceleration due to gravity.
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6) Write the expressions for the electric and magnetic fields, with their corresponding directions, of an electromagnetic wave that has an electric field parallel to the axis and whose amplitude is 300 V/m. Also, this wave has a frequency of 3.0 GHz and travels in the +y direction.
The electric field (E) is along the y-axis and given by E(y, t) = 300 sin(2π(3.0 GHz)t) V/m. The magnetic field (B) is along the x-axis and given by B(y, t) = (300 V/m) / (3.0 x 10^8 m/s) sin(2π(3.0 GHz)t).
The general expression for an electromagnetic wave in free space can be written as:
E(x, t) = E0 sin(kx - ωt + φ)
where:
E(x, t) is the electric field as a function of position (x) and time (t),
E0 is the amplitude of the electric field,
k is the wave number (related to the wavelength λ by k = 2π/λ),
ω is the angular frequency (related to the frequency f by ω = 2πf),
φ is the phase constant.
For the given wave with an electric field parallel to the axis (along the y-axis) and traveling in the +y direction, the expression can be simplified as:
E(y, t) = E0 sin(ωt)
where:
E(y, t) is the electric field as a function of position (y) and time (t),
E0 is the amplitude of the electric field,
ω is the angular frequency (related to the frequency f by ω = 2πf).
In this case, the electric field remains constant in magnitude and direction as it propagates in the +y direction. The amplitude of the electric field is given as 300 V/m, so the expression becomes:
E(y, t) = 300 sin(2π(3.0 GHz)t)
Now let's consider the magnetic field associated with the electromagnetic wave. The magnetic field is perpendicular to the electric field and the direction of wave propagation (perpendicular to the y-axis). Using the right-hand rule, the magnetic field can be determined to be in the +x direction.
The expression for the magnetic field can be written as:
B(y, t) = B0 sin(kx - ωt + φ)
Since the magnetic field is perpendicular to the electric field, its amplitude (B0) is related to the amplitude of the electric field (E0) by the equation B0 = E0/c, where c is the speed of light. In this case, the wave is propagating in free space, so c = 3.0 x 10^8 m/s.
Therefore, the expression for the magnetic field becomes:
B(y, t) = (E0/c) sin(ωt)
Substituting the value of E0 = 300 V/m and c = 3.0 x 10^8 m/s, the expression becomes:
B(y, t) = (300 V/m) / (3.0 x 10^8 m/s) sin(2π(3.0 GHz)t)
To summarize:
- The electric field (E) is along the y-axis and given by E(y, t) = 300 sin(2π(3.0 GHz)t) V/m.
- The magnetic field (B) is along the x-axis and given by B(y, t) = (300 V/m) / (3.0 x 10^8 m/s) sin(2π(3.0 GHz)t).
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A gold wire 5.69 i long and of diameter 0.870 mm
carries a current of 1.35 A For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of
Electrical bazards in bear surgery.
Find the resistance of this wire.
The resistance of the gold wire is 0.235 Ω.
Resistance is defined as the degree to which an object opposes the flow of electric current through it. It is measured in ohms (Ω). Resistance is determined by the ratio of voltage to current. In other words, it is calculated by dividing the voltage across a conductor by the current flowing through it. Ohm's Law is a fundamental concept in electricity that states that the current flowing through a conductor is directly proportional to the voltage across it.
A gold wire with a length of 5.69 cm and a diameter of 0.870 mm is carrying a current of 1.35 A. We need to calculate the resistance of this wire. To do this, we can use the formula for the resistance of a wire:
R = ρ * L / A
In the given context, R represents the resistance of the wire, ρ denotes the resistivity of the material (in this case, gold), L represents the length of the wire, and A denotes the cross-sectional area of the wire. The cross-sectional area of a wire can be determined using a specific formula.
A = π * r²
where r is the radius of the wire, which is half of the diameter given. We can substitute the values given into these formulas:
r = 0.870 / 2 = 0.435 mm = 4.35 × 10⁻⁴ m A = π * (4.35 × 10⁻⁴)² = 5.92 × 10⁻⁷ m² ρ for gold is 2.44 × 10⁻⁸ Ωm L = 5.69 cm = 5.69 × 10⁻² m
Now we can substitute these values into the formula for resistance:R = (2.44 × 10⁻⁸ Ωm) * (5.69 × 10⁻² m) / (5.92 × 10⁻⁷ m²) = 0.235 Ω
Therefore, the resistance of the gold wire is 0.235 Ω.
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Comet C has a gravitational acceleration of 31 m/s?. If its mass is 498 kg, what is the radius of Comet C?
The radius of Comet C is approximately 5.87 x 10^-6 meters, given its mass of 498 kg and gravitational acceleration of 31 m/s².
To calculate the radius of Comet C, we can use the formula for gravitational acceleration:
a = G * (m / r²),
where:
a is the gravitational acceleration,G is the gravitational constant (approximately 6.67430 x 10^-11 m³/(kg·s²)),m is the mass of the comet, andr is the radius of the comet.We can rearrange the formula to solve for r:
r² = G * (m / a).
Substituting the given values:
G = 6.67430 x 10^-11 m³/(kg·s²),
m = 498 kg, and
a = 31 m/s²,
we can calculate the radius:
r² = (6.67430 x 10^-11 m³/(kg·s²)) * (498 kg / 31 m/s²).
r² = 1.0684 x 10^-9 m⁴/(kg·s²) * kg/m².
r² = 3.4448 x 10^-11 m².
Taking the square root of both sides:
r ≈ √(3.4448 x 10^-11 m²).
r ≈ 5.87 x 10^-6 m.
Therefore, the radius of Comet C is approximately 5.87 x 10^-6 meters.
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Given that d=4.3 meters and L=3.5 meters, determine the magnitude of the field at point P in N/C. Assume that P is at the midpoint between the spherical charge and the left edge of the rod.
The magnitude of the electric field at point P is 63 N/C.
The charge of the spherical charge (q_sphere) is 2 μC (2 x 10⁻⁶ C).
The charge of the rod (q_rod) is 5 μC (5 x 10⁻⁶ C).
The distance between the spherical charge and the rod (d) is 2 meters.
Step 1: Calculate the electric field contribution from the spherical charge.
Using the formula:
E_sphere = k * (q_sphere / r²)
Assuming the distance from the spherical charge to point P is r = d/2 = 1 meter:
E_sphere = (9 x 10⁹ N m²/C²) * (2 x 10⁻⁶ C) / (1² m²)
E_sphere = (9 x 10⁹ N m²/C²) * (2 x 10⁻⁶ C) / (1 m²)
E_sphere = 18 N/C
Step 2: Calculate the electric field contribution from the rod.
Using the formula:
E_rod = k * (q_rod / L)
Assuming the length of the rod is L = d/2 = 1 meter:
E_rod = (9 x 10⁹ N m²/C²) * (5 x 10⁻⁶ C) / (1 m)
E_rod = 45 N/C
Step 3: Sum up the contributions from the spherical charge and the rod.
E_total = E_sphere + E_rod
E_total = 18 N/C + 45 N/C
E_total = 63 N/C
So, the magnitude of the electric field at point P would be 63 N/C.
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A sphere of radius R has uniform polarization
P and uniform magnetization M
(not necessarily in the same direction). Calculate the
electromagnetic moment of this configuration.
The electromagnetic moment of a sphere with uniform polarization P and uniform magnetization M can be calculated by considering the electric dipole moment due to polarization and the magnetic dipole moment due to magnetization.
To calculate the electromagnetic moment of the sphere, we need to consider the contributions from both polarization and magnetization. The electric dipole moment due to polarization can be calculated using the formula:
p = 4/3 * π * ε₀ * R³ * P,
where p is the electric dipole moment, ε₀ is the vacuum permittivity, R is the radius of the sphere, and P is the uniform polarization.
The magnetic dipole moment due to magnetization can be calculated using the formula:
m = 4/3 * π * R³ * M,
where m is the magnetic dipole moment and M is the uniform magnetization.
Since the electric and magnetic dipole moments are vectors, the total electromagnetic moment is given by the vector sum of these two moments:
μ = p + m.
Therefore, the electromagnetic moment of the sphere with uniform polarization P and uniform magnetization M is the vector sum of the electric dipole moment due to polarization and the magnetic dipole moment due to magnetization.
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Classify the following statements about Einstein's postulates based on whether they are true or false, True False The speed of light is a constant in all uniformly moving reference frames All reference frames are arbitrary Motion can only be measured relative to one fixed point in the universe. The laws of physics work the same whether the reference frame is at rest or moving at a uniform speed Within a reference frame, it can be experimentally determined whether or not the reference frame is moving The speed of light varies with the speed of the source Answer Bank
According to Einstein's postulates of special relativity, the speed of light in a vacuum is constant and does not depend on the motion of the source or the observer.
This fundamental principle is known as the constancy of the speed of light.
True or False:
1) The speed of light is a constant in all uniformly moving reference frames - True
2) All reference frames are arbitrary - False
3) Motion can only be measured relative to one fixed point in the universe - False
4) The laws of physics work the same whether the reference frame is at rest or moving at a uniform speed - True
5) Within a reference frame, it can be experimentally determined whether or not the reference frame is moving - False
6) The speed of light varies with the speed of the source - False
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2 Magnetic Domain Theory. Answer each of the following questions a) When a bar magnet is broken into two pieces, the two pieces actually become two independent magnets instead of a north-pole magnet and a south-pole magner. Explain this phenomenon b) When a magnet is heated up, it loses it magnetization power. However, when the temperature cools back down, the magnetism power returns (assuming the temperature is lower than the Curie point).
a) When a bar magnet is broken into two pieces, the two pieces become two independent magnets, and not a north-pole magnet and a south-pole magnet. This is because each piece contains its own magnetic domain, which is a region where the atoms are aligned in the same direction. The alignment of atoms in a magnetic domain creates a magnetic field. In a magnet, all the magnetic domains are aligned in the same direction, creating a strong magnetic field.
When a magnet is broken into two pieces, each piece still has its own set of magnetic domains and thus becomes a magnet itself. The new north and south poles of the pieces will depend on the arrangement of the magnetic domains in each piece.
b) When a magnet is heated up, the heat energy causes the atoms in the magnet to vibrate more, which can disrupt the alignment of the magnetic domains. This causes the magnetization power to decrease. However, when the temperature cools back down, the atoms in the magnet stop vibrating as much, and the magnetic domains can re-align, causing the magnetism power to return. This effect is assuming that the temperature is lower than the Curie point, which is the temperature at which a material loses its magnetization permanently.
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A positively-charged object with a mass of 0.191 kg oscillates at the end of a spring, generating ELF (extremely low frequency) radio waves that have a wavelength of 4.40×107 m. The frequency of these radio waves is the same as the frequency at which the object oscillates. What is the spring constant of the spring? Number Units
The spring constant of the spring is approximately 1.90 × 10⁻¹⁷ N/m. This value is obtained by substituting the mass of the object (0.191 kg) and the time period of oscillation (4.35536 × 10¹⁴ s²) into the formula for the spring constant (k = (4π²m) / T²).
According to the information provided, a positively-charged object with a mass of 0.191 kg oscillates at the end of a spring, generating ELF (extremely low frequency) radio waves that have a wavelength of 4.40×10^7 m.
The frequency of these radio waves is the same as the frequency at which the object oscillates. We have to determine the spring constant of the spring. The formula for calculating the spring constant is given as below;k = (4π²m) / T²
Wherek = spring constant
m = mass of the object
T = time period of oscillation
Therefore, first we need to find the time period of oscillation. The formula for time period is given as below;T = 1 / f
Where T = time period
f = frequency
Thus, substituting the given values, we get;
T = 1 / f = 1 / (f (same for radio waves))
Now, to find the spring constant, we substitute the known values of mass and time period into the formula of the spring constant: k = (4π²m) / T²k = (4 x π² x 0.191 kg) / (4.35536 x 10¹⁴ s²) k = 1.90 × 10⁻¹⁷ N/m
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1. (a) At what temperature do the Fahrenheit and Celsius scales have the same numerical value? (b) At what temperature do the Fahrenheit and Kelvin scales have the same numerical value? 1. How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 30 deg C greater than when they were laid? Their 1 original length is 12.5 m. Use a=1.2x10-5 O m
The point at which the Fahrenheit and Celsius scales have the same numerical value is -40°C. The point at which the Fahrenheit and Kelvin scales have the same numerical value is 459.67°F the expansion gap that should be left between the steel railroad rails is 0.0045 m or 4.5 mm.
(a) The point at which the Fahrenheit and Celsius scales have the same numerical value is -40°C. This is because this temperature is equivalent to -40°F. At this temperature, both scales intersect and meet the same numerical value.
(b) The point at which the Fahrenheit and Kelvin scales have the same numerical value is 459.67°F. At this temperature, both scales intersect and meet the same numerical value.
For the second part of the question:
Given that the original length of the steel railroad rails is 12.5m, the maximum temperature rise is 30℃, and the coefficient of linear expansion (a) is 1.2×10⁻⁵/℃.
Therefore, the expansion ΔL can be calculated as:
ΔL = L×a×ΔT
Where L is the original length of the steel railroad rails, a is the coefficient of linear expansion, and ΔT is the temperature rise.
Substituting the given values, we have:
ΔL = 12.5×1.2×10⁻⁵×30
ΔL = 0.0045 m
Therefore, the expansion gap that should be left between the steel railroad rails is 0.0045 m or 4.5 mm. This gap allows the rails to expand without buckling or bending.
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1. Pressure is described as ___ per unit area. a. Flow b. Pounds c. Force d. Inches 2. Pressure is increased when: a. The number of molecules per unit area is decreased Heavier molecules per unit area are introduced b. c. Molecules begin to move faster d. The number of molecules are spread out over a larger ur 3. Atmospheric pressure at sea level is__ _psia? a. 0 b. 2 C. 14.7 d. 27.73
1. Pressure is described as ___ per unit area.
a. Flow
b. Pounds
c. Force
d. Inches
The correct answer is c. Force. Pressure is the force exerted per unit area.
2. Pressure is increased when:
a. The number of molecules per unit area is decreased
b. Heavier molecules per unit area are introduced
c. Molecules begin to move faster
d. The number of molecules are spread out over a larger area
The correct answer is c. Molecules begin to move faster. When molecules move faster, they collide with surfaces more frequently and with greater force, resulting in an increase in pressure.
Atmospheric pressure at sea level is __ psia?
a. 0
b. 2
c. 14.7
d. 27.73
The correct answer is c. 14.7. Atmospheric pressure at sea level is approximately 14.7 pounds per square inch absolute (psia).
Several experiments are performed with light. Which of the following observations is not consistent with the wave model of light? a) The light can travel through a vacuum. b) The speed of the light is less in water than in air. c) The light can exhibit interference patterns when travelling through small openings. d) The beam of light travels in a straight line. e) The light can be simultaneously reflected and transmitted at certain interfaces.
Light has been a matter of extensive research, and experiments have led to various hypotheses regarding the nature of light. The two most notable hypotheses are the wave model and the particle model of light.
These models explain the behavior of light concerning the properties of waves and particles, respectively. Here are the observations for each model:a) Wave model: The light can travel through a vacuum.b) Wave model: The speed of the light is less in water than in air.c) Wave model
e) Wave model: The light can be simultaneously reflected and transmitted at certain interfaces.None of the observations contradicts the wave model of light. In fact, all the above observations are consistent with the wave model of light.The correct answer is d) The beam of light travels in a straight line. This observation is consistent with the particle model of light.
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Two convex thin lenses with focal lengths 12 cm and 18.0 cm aro aligned on a common avis, running left to right, the 12-сm lens being on the left. A distance of 360 сm separates the lenses. An object is located at a distance of 15.0 cm to the left of the 12-сm lens. A Make a sketch of the system of lenses as described above B. Where will the final image appear as measured from the 18-cm bens? Give answer in cm, and use appropriate sign conventions Is the final image real or virtual? D. is the famae upright or inverted? E What is the magnification of the final image?
The magnification is given by: M = v2/v1 = (54 cm)/(60 cm) = 0.9
This means that the image is smaller than the object, by a factor of 0.9.
A. Diagram B. Using the lens formula:
1/f = 1/v - 1/u
For the first lens, with u = -15 cm, f = +12 cm, and v1 is unknown.
Thus,1/12 = 1/v1 + 1/15v1 = 60 cm
For the second lens, with u = 360 cm - 60 cm = +300 cm, f = +18 cm, and v2 is unknown.
Thus,1/18 = 1/v2 - 1/300v2 = 54 cm
Thus, the image is formed at a distance of 54 cm to the right of the second lens, measured from its center, which makes it 54 - 18 = 36 cm to the right of the second lens measured from its right-hand side.
The image is real, as it appears on the opposite side of the lens from the object. It is inverted, since the object is located between the two lenses.
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3. Three polarizing plates whose planes are parallel are centered on a common axis. The directions of the transmission axes relative to the common vertical direction, as shown below. A linearly polarized beam of light with plane of polarization parallel to the vertical reference direction is incident from the left onto the first disk with intensity Ii =10.0 units (arbitrary). If when θ1=20.0∘,θ2=40.0∘, and θ3=60.0∘, then show that the transmitted intensity is about 6.89 units.
The transmitted intensity through the three polarizing plates is approximately 1.296 units.
To determine the transmitted intensity through the three polarizing plates, considering Malus's Law,
I = Ii × cos²(θ)
Where:
I: transmitted intensity
Ii: incident intensity
θ: angle between the transmission axis of the polarizer and the plane of polarization of the incident light.
Given,
Ii = 10.0 units
θ1 = 20.0°
θ2 = 40.0°
θ3 = 60.0°
Calculate the transmitted intensity through each plate:
I₁ = 10.0 × cos²(20.0°)
I₁ ≈ 10.0 × (0.9397)²
I₁ ≈ 8.821 units
I₂ = 8.821 ×cos²(40.0°)
I₂ ≈ 8.821 ×(0.7660)²
I₂ ≈ 5.184 units
I₃ = 5.184 × cos²(60.0°)
I₃ ≈ 5.184 × (0.5000)²
I₃ ≈ 1.296 units
Therefore, the transmitted intensity is 1.296 units.
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We're given a lawnmower with a sound intensity of 0.005 W/m2 at a distance of 3 m. The sound power of the lawnmower works out to be 0.1414 W:
I = P/(4∏r2) --> P = I * (4∏r2)
P = (0.005 W/m2) * (4∏(1.5 m)2)
P = 0.1414 W
Now, you move 20 m away from the lawnmower. What is the intensity level (in dB) from the lawnmower, at this position?
The intensity level from the lawnmower, at a distance of 20 answer: m, is approximately 0.000012 dB.
When we move 20 m away from the lawnmower, we need to calculate the new intensity level at this position. Intensity level is measured in decibels (dB) and can be calculated using the formula:
IL = 10 * log10(I/I0),
where I is the intensity and I0 is the reference intensity (typically 10^(-12) W/m^2).
We can use the inverse square law for sound propagation, which states that the intensity of sound decreases with the square of the distance from the source. The new intensity (I2) can be calculated as follows:
I2 = I1 * (r1^2/r2^2),
where I1 is the initial intensity, r1 is the initial distance, and r2 is the new distance.
In this case, the initial intensity (I1) is 0.005 W/m^2 (given), the initial distance (r1) is 3 m (given), and the new distance (r2) is 20 m (given). Plugging these values into the formula, we get:
I2 = 0.005 * (3^2/20^2)
= 0.0001125 W/m^2.
Convert the new intensity to dB:
Now that we have the new intensity (I2), we can calculate the intensity level (IL) in decibels using the formula mentioned earlier:
IL = 10 * log10(I2/I0).
Since the reference intensity (I0) is 10^(-12) W/m^2, we can substitute the values and calculate the intensity level:
IL = 10 * log10(0.0001125 / 10^(-12))
≈ 0.000012 dB.
Therefore, the intensity level from the lawnmower, at a distance of 20 m, is approximately 0.000012 dB. This value represents a significant decrease in intensity compared to the initial distance of 3 m. It indicates that the sound from the lawnmower becomes much quieter as you move farther away from it.
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(14.9) Atom 1 of mass 38.5 u and atom 2 of mass 40.5 u are both singly ionized with a charge of +e. After being introduced into a mass spectrometer (see the figure below) and accelerated from rest through a potential difference V = 8.09 kV, each ion follows a circular path in a uniform magnetic field of magnitude B = 0.680 T. What is the distance Δx between the points where the ions strike the detector?
The distance Δx between the points where the ions strike the detector is 0.0971 meters. In a mass spectrometer, ions are accelerated by a potential difference and then move in a circular path due to the presence of a magnetic field.
To solve this problem, we can use the equation for the radius of the circular path:
r = (m*v) / (|q| * B)
where m is the mass of the ion, v is its velocity, |q| is the magnitude of the charge, and B is the magnetic field strength. Since the ions are accelerated from rest, we can use the equation for the kinetic energy to find their velocity:
KE = q * V
where KE is the kinetic energy, q is the charge, and V is the potential difference.
Once we have the radius, we can calculate the distance Δx between the two points where the ions strike the detector. Since the ions follow circular paths with the same radius, the distance between the two points is equal to the circumference of the circle, which is given by:
Δx = 2 * π * r
By substituting the given values into the equations and performing the calculations, we find that Δx is approximately 0.0971 meters.
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Suppose you are on another planet and you want to measure its acceleration of gravity so you drop an object from rest. It hits the ground, traveling a distance of 0.8 min 0.5 second and then bounces back up and stops exactly where it started from. a) Please calculate the acceleration of gravity on this planet. b) Taking downward to be positive, how does the ball's average speed compare to the magnitude of its average velocity on the way down? c) Taking the beginning of the motion as the time the ball was dropped, how does its average speed compare to the magnitude of its average velocity on the way up? d) with what speed did the ball hit the ground? e) When distance is divided by time the result is 1.6 m/sec
Given that an object is dropped from rest on another planet and hits the ground, travelling a distance of 0.8 m in 0.5 s and bounces back up and stops exactly where it started from.
Let's find out the acceleration of gravity on this planet. Step-by-step explanation: a) To calculate the acceleration of gravity on this planet, we use the formula d = 1/2 gt².Using this formula, we get0.8 m = 1/2 g (0.5 s)²0.8 m = 0.125 g0.125 g = 0.8 mg = 0.8/0.125g = 6.4 m/s²The acceleration of gravity on this planet is 6.4 m/s².b) Taking downward to be positive, the ball's average speed is equal to its magnitude of average velocity on the way down.
Therefore, the average speed of the ball is equal to the magnitude of its average velocity on the way down.c) The ball's initial speed (when dropped) is zero, so the magnitude of its average velocity on the way up is equal to its final velocity divided by the time taken to stop. Using the formula v = u + gt where v = 0 m/s and u = -6.4 m/s² (negative because the ball is moving up), we get0 = -6.4 m/s² + g*t t = 6.4/gt = √(0.8 m/6.4 m/s²)t = 0.2 seconds.
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(1 p) A beam of light, in air, is incident at an angle of 66° with respect to the surface of a certain liquid in a bucket. If light travels at 2.3 x 108 m/s in such a liquid, what is the angle of refraction of the beam in the liquid?
Given that the beam of light, in air, is incident at an angle of 66° with respect to the surface of a certain liquid in a bucket, and the light travels at 2.3 x 108 m/s in such a liquid, we need to calculate the angle of refraction of the beam in the liquid.
We can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the velocities of light in the two media. Mathematically, it can be expressed as:
n₁sinθ₁ = n₂sinθ₂
where n₁ and n₂ are the refractive indices of the first and second medium respectively; θ₁ and θ₂ are the angles of incidence and refraction respectively.
The refractive index of air is 1 and that of the given liquid is not provided, so we can use the formula:
n = c/v
where n is the refractive index, c is the speed of light in vacuum (3 x 108 m/s), and v is the speed of light in the given medium (2.3 x 108 m/s in this case). Therefore, the refractive index of the liquid is:
n = c/v = 3 x 10⁸ / 2.3 x 10⁸ = 1.3043 (approximately)
Now, applying Snell's law, we have:
1 × sin 66° = 1.3043 × sin θ₂
⇒ sin θ₂ = 0.8165
Therefore, the angle of refraction of the beam in the liquid is approximately 54.2°.
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Create a dictionary of physical terms and write by hand from a physics textbook (Baryakhtar) the definitions of the following concepts and some formulas:
Electric charge + [formula demonstrating the discreteness of electric charge]
Electrification
Electric field
Electric field lines of force
Law of conservation of electric charge
Coulomb's law + [Coulomb's law formula]
Electric current
Conductors
Dielectrics
Electrical diagram + [redraw the symbols of the main elements of the electrical circuit]
Amperage + [amperage formula]
Electric voltage + [voltage formula]
Electrical resistance + [resistance formula]
Volt-ampere characteristic of the conductor
Specific resistance of the substance + [formula of the specific resistance of the substance]
Rewrite the basic formulas for serial connection
Rewrite the basic formulas for parallel connection
Electric current power + [electric current power formula]
Joule-Lenz law + [formula for the Joule-Lenz law]
Electric current in metals
Electrolytic dissociation
Electric current in electrolytes
Electrolytes
Electrolysis
Faraday's first law + [Faraday's first law formula]
Galvanostegia
Ionization
Electric current in gases
Write SI units for charge, current, voltage, resistance, work, power.
Study the infographic on p. 218-219.
Solve problems:
Two resistors are connected in series in the circuit. The resistance of the first is 60 ohms; a current of 0.1 A flows through the second. What will be the resistance of the second resistor if the battery voltage is 9 V?
Two bulbs are connected in parallel. The voltage and current in the first bulb are 50 V and 0.5 A. What will be the total resistance of the circuit if the current in the second bulb is 2 A?
Calculate the current strength and the work it performs in 20 minutes, if during this time 1800 K of charge passes through the device at a voltage of 220 V.
This is a dictionary of physical terms and formulas related to electricity, including definitions and problem-solving examples on electric current, voltage, and resistance. The resistance of the 2nd resistor is 54 [tex]\Omega[/tex], the total resistance of the circuit is 25 [tex]\Omega[/tex] and the current strength is 1.5 A, and the work is 198000 J
A dictionary of physical terms comprises Electric charge, Electrification, Electric field, Electric field lines of force, Law of conservation of electric charge, Coulomb's law, Electric current, Conductors, Dielectrics, Electrical diagram, Amperage, Electric voltage, Electrical resistance, Volt-ampere characteristic of the conductor, Specific resistance of the substance, Rewriting of the basic formulas for serial connection, Rewriting of the basic formulas for parallel connection, Electric current power, Joule-Lenz law, Electric current in metals, Electrolytic dissociation, Electric current in electrolytes, Electrolytes, Electrolysis, Faraday's first law, Galvanostegia, Ionization, Electric current in gases, and SI units for a charge, current, voltage, resistance, work, and power. A battery voltage of 9 V flows through two resistors connected in a series in the circuit. The resistance of the first resistor is 60 ohms, and a current of 0.1 A flows through the second. The resistance of the second resistor will be 54 ohms. Two bulbs are connected in parallel, and the voltage and current in the first bulb are 50 V and 0.5 A. The total resistance of the circuit will be 25 ohms if the current in the second bulb is 2 A. If 1800 K of charge passes through the device at a voltage of 220 V in 20 minutes, the current strength and the work it performs can be calculated, and the current strength is 1.5 A, and the work is 198000 J (Joules). Hence, this is about a dictionary of physical terms along with some formulas and definitions along with problem-solving on electric current, electric voltage, and electrical resistance in a detailed manner.For more questions on electric current
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Calculate heat loss by metal and heat gained by water with the
following information.
Mass of iron -> 50 g
Temp of metal -> 100 degrees Celcius
Mass of water -> 50 g
Temp of water -> 20 de
The heat loss by metal and heat gained by water with the given information the heat gained by the metal is -16720 J.
We can use the following calculation to determine the heat loss by the metal and the heat gained by the water:
Q = m * c * ΔT
Here, it is given:
m1 = 50 g
T1 = 100 °C
c1 = 0.45 J/g°C
m2 = 50 g
T2 = 20 °C
c2 = 4.18 J/g°C
Now, the heat loss:
ΔT1 = T1 - T2
ΔT1 = 100 °C - 20 °C = 80 °C
Q1 = m1 * c1 * ΔT1
Q1 = 50 g * 0.45 J/g°C * 80 °C
Now, heat gain,
ΔT2 = T2 - T1
ΔT2 = 20 °C - 100 °C = -80 °C
Q2 = m2 * c2 * ΔT2
Q2 = 50 g * 4.18 J/g°C * (-80 °C)
Q1 = 50 g * 0.45 J/g°C * 80 °C
Q1 = 1800 J
Q2 = 50 g * 4.18 J/g°C * (-80 °C)
Q2 = -16720 J
Thus, as Q2 has a negative value, the water is losing heat.
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A circuit operating at 90 Hz and contains only two circuit elements, but it is not known if they are L, R, or C. A maximum voltage of 175 V is applied by the source. If the maximum current in the circuit is 13.6 A and lags the voltage by 37 ∘
, a. Draw a phashor diagram of this circuit b. What two circuit elements are connected? Explain c. Calculate the values of the two circuit elements.
Resistance (R) = 12.87 Ω
Inductance (L) = 35 mH (or 0.000035 H)
a. Phasor diagram of the circuit is given below:b. The two circuit elements are connected are inductance (L) and resistance (R).
In a purely inductive circuit, voltage and current are out of phase with each other by 90°. In a purely resistive circuit, voltage and current are in phase with each other. Hence, by comparing the phase difference between voltage and current, we can determine that the circuit contains inductance (L) and resistance (R).
c. We know that;
Maximum voltage (V) = 175 VMaximum current (I) = 13.6
APhase angle (θ) = 37°
We can find out the Impedance (Z) of the circuit by using the below relation;
Impedance (Z) = V / IZ = 175 / 13.6Z = 12.868 Ω
Now, we can find out the values of resistance (R) and inductance (L) using the below relations;
Z = R + XL
Here, XL = 2πfL
Where f = 90 Hz
Therefore,
XL = 2π × 90 × LXL = 565.49 LΩ
Z = R + XL12.868 Ω = R + 565.49 LΩ
Maximum current (I) = 13.6 A,
so we can calculate the maximum value of R and L using the below relations;
V = IZ175 = 13.6 × R
Max R = 175 / 13.6
Max R = 12.87 Ω
We can calculate L by substituting the value of R
Max L = (12.868 − 12.87) / 565.49
Max L = 0.000035 H = 35 mH
Therefore, the two circuit elements are;
Resistance (R) = 12.87 Ω
Inductance (L) = 35 mH (or 0.000035 H)
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: The position of a partide moving along the x axis is given in centimeters by-7.00+ 2.50e, where it is in seconds. Consider the time interval 2.00 tot-3.00 s (ndicate the direction with the sign of your answer.) (a) Calculate the average velocity. cm/s (b) Calculate the instantaneous velocity at t-2.00 s cm/s (c) Calculate the instantaneous velocity at t-3.00 s om/s (d) Calculate the instantaneous velocity at r-2.50 s cm/s (e) Calculate the instantaneous velocity when the particle is midway between its positions at -2.00 and 3.00 cm/s (f) Graph x versus t and indicate your answers graphically.
(a) The average velocity of the particle during the time interval from 2.00 to 3.00 seconds is -2.50 cm/s.
(b) The instantaneous velocity at t = 2.00 seconds is -2.50 cm/s.
(c) The instantaneous velocity at t = 3.00 seconds is -2.50 cm/s.
(d) The instantaneous velocity at t = 2.50 seconds is -2.50 cm/s.
(e) The instantaneous velocity when the particle is midway between its positions at -2.00 and 3.00 seconds is -2.50 cm/s.
(f) The graph of x versus t would show a linear relationship with a downward slope of -2.50 cm/s.
The given equation for the position of the particle along the x-axis is -7.00 + 2.50e, where t represents time in seconds. In this equation, the term -7.00 represents the initial position of the particle at t = 0 seconds, and 2.50e represents the displacement or change in position with respect to time.
(a) To calculate the average velocity, we need to find the total displacement of the particle during the given time interval and divide it by the duration of the interval.
In this case, the displacement is given by the difference between the positions at t = 3.00 seconds and t = 2.00 seconds, which is (2.50e) at t = 3.00 seconds minus (2.50e) at t = 2.00 seconds. Simplifying this expression, we get -2.50 cm/s as the average velocity.
(b) The instantaneous velocity at t = 2.00 seconds can be found by taking the derivative of the position equation with respect to time and evaluating it at t = 2.00 seconds. The derivative of -7.00 + 2.50e with respect to t is simply 2.50e. Substituting t = 2.00 seconds into this expression, we get -2.50 cm/s as the instantaneous velocity.
(c) Similarly, to find the instantaneous velocity at t = 3.00 seconds, we evaluate the derivative 2.50e at t = 3.00 seconds, which also gives us -2.50 cm/s.
(d) The instantaneous velocity at t = 2.50 seconds can be determined in the same way, by evaluating the derivative 2.50e at t = 2.50 seconds, resulting in -2.50 cm/s.
(e) When the particle is midway between its positions at -2.00 and 3.00 seconds, the time is 2.00 + (3.00 - 2.00)/2 = 2.50 seconds. Therefore, the instantaneous velocity at this time is also -2.50 cm/s.
(f) The graph of x versus t would be a straight line with a slope of 2.50 cm/s, indicating a constant velocity of -2.50 cm/s throughout the given time interval.
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What is the best possible coefficient of performance COPret for a refrigerator that cools an environment at -13.0°C and exhausts heat to another environment at 39.0°C? COPrel= How much work W would this ideal refrigerator do to transfer 3.125 x 10 J of heat from the cold environment? W = What would be the cost of doing this work if it costs 10.5¢ per 3.60 × 106 J (a kilowatt-hour)? cost of heat transfer: How many joules of heat Qu would be transferred into the warm environment?
The best possible coefficient of performance (COPret) for the given temperatures is approximately 5.0. The work done by the refrigerator is calculated to be 6.25 x 10 J. The cost of performing this work is approximately 0.0182¢. Finally, the amount of heat transferred into the warm environment is determined to be 9.375 x 10.
The coefficient of performance (COP) of a refrigerator is a measure of its efficiency and is defined as the ratio of the amount of heat transferred from the cold environment to the work done by the refrigerator. For an ideal refrigerator, the COP can be determined using the formula:
COPret = Qc / W
where Qc is the amount of heat transferred from the cold environment and W is the work done by the refrigerator.
To find the best possible COPret for the given temperatures, we need to use the Carnot refrigerator model, which assumes that the refrigerator operates in a reversible cycle. The Carnot COP (COPrel) can be calculated using the formula:
COPrel = Th / (Th - Tc)
where Th is the absolute temperature of the hot environment and Tc is the absolute temperature of the cold environment.
Converting the given temperatures to Kelvin, we have:
Th = 39.0°C + 273.15 = 312.15 K
Tc = -13.0°C + 273.15 = 260.15 K
Substituting these values into the equation, we can calculate the COPrel:
COPrel = 312.15 K / (312.15 K - 260.15 K) ≈ 5.0
Now, we can use the COPrel value to determine the work done by the refrigerator. Rearranging the COPret formula, we have:
W = Qc / COPret
Given that Qc = 3.125 x 10 J, we can calculate the work done:
W = (3.125 x 10 J) / 5.0 = 6.25 x 10 J
Next, we can calculate the cost of doing this work, considering the given cost of 10.5¢ per 3.60 × 10^6 J (a kilowatt-hour). First, we convert the work from joules to kilowatt-hours:
W_kWh = (6.25 x 10 J) / (3.60 × 10^6 J/kWh) ≈ 0.0017361 kWh
To calculate the cost, we use the conversion rate:
Cost = (0.0017361 kWh) × (10.5¢ / 1 kWh) ≈ 0.01823¢ ≈ 0.0182¢
Finally, we need to determine the amount of heat transferred into the warm environment (Qw). For an ideal refrigerator, the total heat transferred is the sum of the heat transferred to the cold environment and the work done:
Qw = Qc + W = (3.125 x 10 J) + (6.25 x 10 J) = 9.375 x 10 J
In summary, the best possible coefficient of performance (COPret) for the given temperatures is approximately 5.0. The work done by the refrigerator is calculated to be 6.25 x 10 J. The cost of performing this work is approximately 0.0182¢. Finally, the amount of heat transferred into the warm environment is determined to be 9.375 x 10.
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Destructive interference of two superimposed waves requires the waves to travel in opposite directions. Select one: True False
The given statement, "Destructive interference of two superimposed waves requires the waves to travel in opposite directions" is false because destructive interference of two superimposed waves requires the waves to be traveling in the same direction and having a phase difference of π or an odd multiple of π.
In destructive interference, the two waves will have a phase difference of either an odd multiple of π or an odd multiple of 180 degrees. When the phase difference is an odd multiple of π, it results in a complete cancellation of the two waves in the region where they are superimposed and the resultant wave has zero amplitude. In constructive interference, the two waves will have a phase difference of either an even multiple of π or an even multiple of 180 degrees. When the phase difference is an even multiple of π, it results in a reinforcement of the two waves in the region where they are superimposed and the resultant wave has maximum amplitude.
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A 688.78 mm long aluminum wire with a diameter of 41.4 mm changes temperature from 131.6 C to 253.3 C. Calculate the change in length of the wire due to the temperature change. Report your answer in millimeters rounded to 3 decimal places with units.
We know that the coefficient of linear expansion of aluminum, α = 23.1 x 10-6 K-1 Hence,∆L = αL∆T= 23.1 × 10-6 × 688.78 × (253.3 − 131.6)= 4.655 mmThus, the change in length of the wire due to the temperature change is 4.655 mm (rounded to 3 decimal places with u
The length change of an aluminum wire with a diameter of 41.4 mm and 688.78 mm length from a temperature change from 131.6 C to 253.3 C is 4.655 mm. The formula that is used to calculate the change in length of the wire due to temperature change is:∆L
= αL∆T
where, ∆L is the change in length L is the original length of the wireα is the coefficient of linear expansion of the material of the wire∆T is the change in temperature From the provided data, we know the following:Length of the aluminum wire
= 688.78 mm Diameter of the aluminum wire
= 41.4 mm Radius of the aluminum wire
= Diameter/2
= 41.4/2
= 20.7 mm Initial temperature of the aluminum wire
= 131.6 C Final temperature of the aluminum wire
= 253.3 C
We first need to find the coefficient of linear expansion of aluminum. From the formula,α
= ∆L/L∆T We know that the change in length, ∆L
= ?L = 688.78 mm (given)We know that the initial temperature, T1
= 131.6 C
We know that the final temperature, T2
= 253.3 C.We know that the coefficient of linear expansion of aluminum, α
= 23.1 x 10-6 K-1 Hence,∆L
= αL∆T
= 23.1 × 10-6 × 688.78 × (253.3 − 131.6)
= 4.655 mm Thus, the change in length of the wire due to the temperature change is 4.655 mm (rounded to 3 decimal places with units).
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A point charge Q₁ = +64 μC is 88 cm away from another point charge Q₂ = -32 HC. The direction of the electric force acting on Q₁ is:
A) Pushing Q1 directly away from Q2
B) some other direction
C) Pushing Q1 directly towards Q2
A point charge Q₁ = +64 μC is 88 cm away from another point charge Q₂ = -32 HC. The direction of the electric force acting on Q₁ is pushing Q1 directly towards Q2 which is in option C.
The formula for the magnitude of the electric force (F) between two point charges is given by:
F = (k × |Q₁ × Q₂|) / r²
Where:
F is the magnitude of the electric force
k is the Coulomb's constant (k ≈ 8.99 x 1[tex]0^9[/tex] N m²/C²)
Q₁ and Q₂ are the magnitudes of the charges
r is the distance between the charges
In this case, Q₁ = +64 μC and Q₂ = -32 μC, and the distance between them is 88 cm = 0.88 m.
Plugging in the values into Coulomb's law:
F = (8.99 x 1[tex]0^9[/tex] N m²/C² × |(+64 μC) × (-32 μC)|) / (0.88 m)²
Calculating the value:
F ≈ (8.99 x 1[tex]0^9[/tex] N m²/C² * (64 x 10^-6 C) * (32 x 1[tex]0^-^6[/tex] C)) / (0.88 m)²
F ≈ (8.99 x 1[tex]0^9[/tex] N m²/C² ×2.048 x 1[tex]0^-^6[/tex] C²) / 0.7744 m²
F ≈ 23.84 N
Now, after analyzing the sign of the force. Since Q₁ is positive (+) and Q₂ is negative (-), the charges have opposite signs. The electric force between opposite charges is attractive, which means it acts towards each other.
Therefore, the electric force acting on Q₁ is pushing it directly towards Q₂.
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Option (C) is correct, Pushing Q1 directly towards Q2
The electric force acting on Q₁ will be directed towards Q₂ which is 88 cm away from Q₁. The correct option is (C) Pushing Q1 directly towards Q2.
Electric force is the force between two charged particles. It is a fundamental force that exists between charged objects. Like gravity, the electric force between two particles is an attractive force that is directly proportional to the product of the charges on the two particles and inversely proportional to the square of the distance between them.In the given problem, there are two charges: Q₁ = +64 μC and Q₂ = -32 HC and the distance between them is 88 cm. Now, we have to find the direction of the electric force acting on Q₁. Since the charges are of opposite sign, they will attract each other. The force on Q₁ due to Q₂ will be directed towards Q₂. The direction of the electric force acting on Q₁ is:Pushing Q₁ directly towards Q₂.
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Heat is produced within a cylindrical cable with a radius of 0.60 m and a length of 3 m with a heat conductivity of 85 W/m K. The amount of heat produced per unit volume and per unit time is given as Q (W/m3.s) = 4x10-3 T0.5 where T is the temperature (K). The surface temperature of the sphere is 120 °C. a) Construct an energy balance within the cylindrical cable. b) Solve the energy balance with MATLAB to obtain the temperature profile within the cylindrical cable by appropriate assumptions
The energy balance equation for the cylindrical cable can be constructed by considering the heat generation, heat conduction, and heat transfer at the boundaries.
a) Energy balance within the cylindrical cable: The energy balance equation for the cylindrical cable can be constructed by considering the heat generation, heat conduction, and heat transfer at the boundaries. The heat generated per unit volume is given by Q (W/m3.s) = 4x10-3 T0.5, where T is the temperature. The heat conduction within the cable can be described by Fourier's law of heat conduction. The energy balance equation can be written as the sum of the rate of heat generation and the rate of heat conduction, which should be equal to zero for steady-state conditions. The equation can be solved to determine the temperature profile within the cable.
b) Solving the energy balance with MATLAB: To obtain the temperature profile within the cylindrical cable, MATLAB can be used to numerically solve the energy balance equation. The equation involves a second-order partial differential equation, which can be discretized using appropriate numerical methods like finite difference or finite element methods. By discretizing the cable into small segments and solving the equations iteratively, the temperature distribution can be obtained. Assumptions such as uniform heat generation, isotropic heat conductivity, and steady-state conditions can be made to simplify the problem. MATLAB provides built-in functions and tools for solving partial differential equations, making it suitable for this type of analysis. By implementing the appropriate numerical method and applying boundary conditions, the temperature profile within the cylindrical cable can be calculated using MATLAB.
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Superman must stop a 190-km/h train in 200 m to keep it from hitting a stalled car on the tracks Part A If the train's mass is 3.7x105 kg, how much force must he exert (find the magnitude)? Express your answer using two significant figures.
The force required to stop the train is 2.93 × 10⁶ N (to two significant figures).
Given that Superman must stop a 190-km/h train in 200 m to keep it from hitting a stalled car on the tracks. The train's mass is 3.7 × 10⁵ kg.
To calculate the force, we use the formula:
F = ma
Where F is the force required to stop the train, m is the mass of the train, and a is the acceleration of the train.
So, first, we need to calculate the acceleration of the train. To calculate acceleration, we use the formula:
v² = u² + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
The initial velocity of the train is 190 km/h = 52.8 m/s (since 1 km/h = 1000 m/3600 s)
The final velocity of the train is 0 m/s (since Superman stops the train)
The distance traveled by the train is 200 m.
So, v² = u² + 2as ⇒ (0)² = (52.8)² + 2a(200) ⇒ a = -7.92 m/s² (the negative sign indicates that the train is decelerating)
Now, we can calculate the force:
F = ma = 3.7 × 10⁵ kg × 7.92 m/s² = 2.93 × 10⁶ N
Therefore, the force required to stop the train is 2.93 × 10⁶ N (to two significant figures).
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(a) A defibrillator connected to a patient passes 15.0 A of
current through the torso for 0.0700 s. How much charge moves? C
(b) How many electrons pass through the wires connected to the
patient? ele
1.05 Coulombs of charge moves through the torso and approximately 6.54 × 10^18 electrons pass through the wires connected to the patient.
(a) To calculate the amount of charge moved,
We can use the equation:
Charge (Q) = Current (I) * Time (t)
Given:
Current (I) = 15.0 A
Time (t) = 0.0700 s
Substituting the values into the equation:
Q = 15.0 A * 0.0700 s
Q = 1.05 C
Therefore, 1.05 Coulombs of charge moves.
(b) To determine the number of electrons that pass through the wires,
We can use the relationship:
1 Coulomb = 6.242 × 10^18 electrons
Given:
Charge (Q) = 1.05 C
Substituting the value into the equation:
Number of electrons = 1.05 C * 6.242 × 10^18 electrons/Coulomb
Number of electrons ≈ 6.54 × 10^18 electrons
Therefore, approximately 6.54 × 10^18 electrons pass through the wires connected to the patient.
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