The electric flux through the inner Gaussian surface is equal to the electric flux through the outer Gaussian surface. This is because the electric flux through a closed surface depends only on the total charge enclosed by that surface, not on the size or shape of the surface.
The electric flux through a closed surface is given by the equation:
Φ = E * A * cos(θ)
where Φ is the electric flux, E is the electric field, A is the area of the surface, and θ is the angle between the electric field and the normal to the surface.
In this case, we have a point charge located at the center of the Gaussian surfaces. The electric field due to a point charge decreases with distance according to the inverse square law, so the magnitude of the electric field will be weaker for the larger Gaussian surface compared to the smaller one.
However, the areas of the two Gaussian surfaces are different. The outer surface has a larger area than the inner surface because it has a larger radius. The decrease in the magnitude of the electric field is compensated by the increase in the area, resulting in the same electric flux through both surfaces.
To visualize this, imagine the point charge as the source of electric field lines radiating outward. The field lines passing through the inner surface will be more concentrated due to the smaller area, while the field lines passing through the outer surface will be more spread out due to the larger area. However, the total number of field lines passing through each surface will be the same since they originate from the same point charge.
Therefore, the electric flux through the inner Gaussian surface is equal to the electric flux through the outer Gaussian surface.
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If the mass of the Sun were to decrease to half its current value, what orbital velocity would the Earth need to travel at to maintain its current distance to the Sun?
To maintain the same orbital period, the Earth would need to increase its orbital velocity.
If the mass of the Sun were to decrease to half its current value, the Earth would need to increase its orbital velocity to maintain its current distance from the Sun.
The orbital velocity of a planet around a star depends on the mass of the star and the distance between them. According to Kepler's third law, the square of the orbital period (T) is directly proportional to the cube of the semi-major axis (r) of the orbit. Mathematically, T^2 ∝ r^3.
Since the Earth's distance from the Sun (r) would remain the same, the orbital period (T) would also remain the same. Therefore, to maintain the same orbital period, the Earth would need to increase its orbital velocity.
The relationship between orbital velocity (V) and the mass of the central star (M) is given by the equation V ∝ √(M/r), where V is the orbital velocity. If the mass of the Sun decreases to half its current value, the Earth would need to increase its orbital velocity by a factor of √2 to compensate for the decreased gravitational pull and maintain its current distance from the Sun.
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A radar pulse travels at the speed of light to the moon with the reflected pulse returning to earth 2.59 seconds later. How far is the moon from the radar transmitter on earth? Express your answer in km.
If a radar pulse travels at the speed of light to the moon with the reflected pulse returning to earth 2.59 seconds later. The distance between the moon and the radar transmitter on Earth is estimated to be approximately 776,951.21 km.
The distance between the moon and the radar transmitter on Earth can be calculated using the formula:
Distance = Speed × Time
Where:
Distance is the distance between Earth and Moon
Speed is the speed of light, approximately 299,792,458 meters per second (m/s)
Time is the time taken by the radar pulse to reach the moon and come back, which is 2.59 seconds.
Using the given values in the formula, we can calculate the distance as follows:
Distance = 299,792,458 m/s × 2.59 s
Distance = 776,951,212.22 meters
To express the answer in kilometers, we divide the distance by 1000:
Distance in km = 776,951,212.22 m / 1000
Distance in km = 776,951.21 km
Therefore, the distance between the moon and the radar transmitter on Earth is approximately 776,951.21 km.
By multiplying the speed of light by the time taken, we obtain the distance in meters. Converting the result to kilometers, we divide by 1000. The final result is approximately 776,951.21 km, indicating the distance between the moon and the radar transmitter on Earth.
The distance between the moon and the radar transmitter on Earth is estimated to be approximately 776,951.21 km. The calculation involves multiplying the speed of light by the time taken for the radar pulse to travel to the moon and back, and converting the result from meters to kilometers.
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Given a DTMF transmission signal y = A sin(2pif1t) + Azsin(2pif2t) where the frequency of the signals will be decided on the number typed between 0-9. The amplitude of the first signal is A1 = 120 and the second one Az 180. = 1. Write a MATLAB code to generate appropriate signal for the dialed number-5 and draw it in Time & Frequency domain. (10 points) 2. A Gaussian noise with mean 0 and variance 2 is added to the signal. Write a MATLAB code to plot the Received signal in Time & Frequency domain. (10 points) 3. Show all 4 figures in one graph. (5 points)
The original signal in the time domain, the received signal with noise in the time domain, the frequency spectrum of the original signal, and the frequency spectrum of the received signal with noise.
Here's a MATLAB code that generates the DTMF transmission signal for the dialed number 5, adds Gaussian noise, and plots the resulting signal in both the time and frequency domains.
% Constants
A1 = 120; % Amplitude of the first signal
Az = 180; % Amplitude of the second signal
f1 = 697; % Frequency of the first signal
f2 = 1336; % Frequency of the second signal
variance = 2; % Variance of the Gaussian noise
% Time parameters
Fs = 8000; % Sampling frequency
duration = 1; % Duration of the signal (in seconds)
t = 0:1/Fs:duration-1/Fs; % Time vector
% Generate DTMF transmission signal for number 5
y = A1×sin(2×pi×f1×t) + Az×sin(2×pi×f2×t);
% Add Gaussian noise
noise = √(variance)×randn(size(y));
y(noisy) = y + noise;
% Plot the original and received signals in the time domain
subplot(2,2,1);
plot(t, y);
title('Original Signal (Number 5)');
xlabel('Time (s)');
ylabel('Amplitude');
subplot(2,2,2);
plot(t, y(noisy));
title('Received Signal with Noise (Number 5)');
xlabel('Time (s)');
ylabel('Amplitude');
% Compute and plot the frequency spectrum of the original signal
subplot(2,2,3);
Y = fft(y);
frequencies = linespace(0, Fs, length(Y));
plot(frequencies, abs(Y));
title('Frequency Spectrum of Original Signal');
xlabel('Frequency (Hz)');
ylabel('Magnitude');
% Compute and plot the frequency spectrum of the received signal
subplot(2,2,4);
Y(noisy) = fft(y(noisy));
plot(frequencies, abs(Y(noisy)));
title('Frequency Spectrum of Received Signal with Noise');
xlabel('Frequency (Hz)');
ylabel('Magnitude');
% Adjust subplot spacing
sgtitle('DTMF Signal and Received Signal with Noise');
This code will generate four figures: the original signal in the time domain, the received signal with noise in the time domain, the frequency spectrum of the original signal, and the frequency spectrum of the received signal with noise. All four figures will be displayed in a single graph using subplots.
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What is conventional current and what makes different with electron current?
I am currently confuse because usually conventional current is associated as "the movement of proton" while what I know proton is hard to move and once it moves it occurs nuclear disaster. Thus, the term of "electric current" is defined as the flow of "electrons" or "protons" (which is conventional current)?
Conventional current is the flow of positive charges from higher potential to lower potential. It is the opposite direction of electron current, which refers to the movement of negatively charged electrons.
Conventional current is a concept used in electrical circuits that assumes the flow of positive charges from higher potential (positive terminal) to lower potential (negative terminal). This convention was established before the discovery of electrons, so it represents the direction of current flow in a circuit. It simplifies calculations and aligns with the direction of current in conventional circuits.
On the other hand, electron current refers to the actual movement of negatively charged electrons. Electrons flow from lower potential (negative terminal) to higher potential (positive terminal) in a circuit. This is the opposite direction of conventional current.
While the flow of electrons constitutes the physical current in a conductor, conventional current provides a simplified representation that facilitates analysis and understanding of circuit behavior. It is important to keep in mind the distinction between the two when interpreting circuit diagrams or discussing the direction of current flow.
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A lightwave from a star has a frequency of 6.67 x 1014 Hz and a wavelength of 4.50 x 10-7 m. The star is 4.00 x 1016 m away from the earth. Calculate the velocity of the light of the star to reach the Earth. Choose the correct answer but submit your calculation. A. 3.33 x 108 m/s OB. 300 x 108 m/s OC. 3.00 x x 108 m/s O D.0.30 x 108 m/s QUESTION 20 A light ray strikes a reflective plane surface at an angle of 56° with the surface. What would the respective angles of incidence and reflection be? A. 34° 44° B. 34° 34° C. 30° 34° D. 38 34°
The answer is: B. 34° 34°. The velocity of light from the star to reach the Earth is approximately 3.00 x [tex]10^8[/tex] m/s.
The velocity of light in a vacuum is a fundamental constant, denoted by the symbol "c." Its value is approximately 3.00 x [tex]10^8[/tex] m/s.
Given that the wavelength (λ) and frequency (f) of the lightwave from the star are provided, we can use the equation:
c = λf
Substituting the given values:
3.00 x [tex]10^8[/tex] m/s = (4.50 x [tex]10^(-7)[/tex] m)(6.67 x [tex]10^{14}[/tex] Hz)
Simplifying the right side of the equation:
3.00 x [tex]10^8[/tex]m/s = 3.00 x [tex]10^8[/tex]m/s
The equation holds true, confirming that the velocity of light is indeed 3.00 x [tex]10^8[/tex]m/s.
Therefore, the velocity of light from the star to reach the Earth is approximately 3.00 x [tex]10^8[/tex] m/s.
For the second question, the angles of incidence and reflection are equal for a ray of light striking a reflective plane surface. Since the angle of incidence is given as 56°, the respective angles of incidence and reflection would both be 56°.
Therefore, the answer is: B. 34° 34°
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A wire loop with diameter D = 2 cm and resistance R = 10−4 Ω is inside a solenoid with a winding density of n = 200 m−1 . The current in the solenoid is gradually increased from 0 to I2 = 5 A within 10 ms. Draw a sketch showing the direction of the magnetic field and the induced emf in the loop.
The current in the solenoid creates a magnetic field inside the solenoid. The magnetic field is proportional to the current, so as the current increases, the magnetic field increases.
The wire loop is a conductor, and a changing magnetic field can induce a current in a conductor. In this case, the changing magnetic field in the solenoid induces a current in the wire loop. The direction of the induced current is such that it opposes the change in the magnetic field. This is known as Lenz's law. The induced current in the wire loop creates a magnetic field of its own. This magnetic field opposes the magnetic field from the solenoid. As the current in the solenoid increases, the induced current in the wire loop increases. This creates a stronger and stronger opposing magnetic field. Eventually, the opposing magnetic field becomes strong enough to stop the increase in current in the solenoid.
The time it takes for the current in the solenoid to reach 5 A is 10 ms. The induced current in the wire loop will also reach 5 A at this time. The magnitude of the induced current will be equal to the current in the solenoid. The direction of the induced current will be opposite to the current in the solenoid.
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3. Use normal vectors to determine the intersection, if any, for each of the following groups of three planes. Give a geometric interpretation in each case and the number of solutions for the correspo
In normal vectors, the system of equations has no solution because: 0 ≠ 3. The system of equations has no solution because 0 ≠ 1. The system of equations has no solution because: 0 ≠ 17. The system of equations has no solution because: 0 ≠ 4. x₁ = 2x2 = 1x3 = -3. The system of equations has no solution because 0 ≠ 1.
x₁ + 2x₂ + 3x₃ = -42x₁ + 4x₂ + 6x₃ = 73x₁ + 6x₂+ 9x₃ = -3
Simplify the system:
x₁ + 2x₂ + 3x₃ = -42x₁ + 4x₂ + 6x₃ = 7x₁ + 2x₂+ 3x₃ = -1
From 1-th equation we find x1 by other variables
x₁ = -2x₂ - 3x₃ - 42x₁ + 4x₂ + 6x₃ = 7x₁ + 2x₂ + 3x₃ = -1
In 2, 3-th equation substitute x1
x₁ = -2x₂ - 3x₃ - 42(-2x₂ - 3x₃ - 4) + 4x₂ + 6x₃ = 7(-2x₂ - 3x₃ - 4) + 2x₂ + 3x₃ = -1
After simplification, we get:
x₁ = -2x₂ - 3x₃ - 40 = 150 = 3
Answer:
The system of equations has no solution because: 0 ≠ 3.
B)Solve the system of equations:
x₁ + 2x₂ + 3x₃ = -42x₁ + 4x₂ + 6x₃ = 73x₁ + 6x₂+ 9x₃ = 5
From 1-th equation we find x1 by other variables
x₁ = -2x₂ - 3x₃ - 42x₁ + 4x₂ + 6x₃ = 73x₁ + 6x₂+ 9x₃ = 5
In 2, 3-th equation substitute x₁
x₁ = -2x₂ - 3x₃ - 42(-2x₂ - 3x₃ - 4) + 4x₂ + 6x₃= 73(-2x₂ - 3x₃ - 4) + 6x₂ + 9x₃ = 5
After simplification,:
x1 = -2x₂ - 3x₃ - 40 = 150 = 17
Answer:
The system of equations has no solution because: 0 ≠ 17
C)Solve the system of equations:
x₁ + 2x₂ + x₃ = -22x₁ + 4x₂ + 2x₃ = 43x₁ + 6x₂+ 3x₃ = -6
Simplify the system:
x1 + 2x₂ + x₃ = -2x₁ + 2x₂ + x₃ = 2x₁ + 2x₂ + x₃= -2
From 1-th equation, we find x₁ by other ₁
x₁ = -2x₂ - x₃ - 2x₁ + 2x₂ + x₃ = 2x₁ + 2x₂ + x₃= -2
In 2, 3-th equation substitute x₁
x₁ = -2x₂ - x₃ - 2(-2x₂ - x₃ - 2) + 2x₂ + x₃ = 2(-2x₂ - x₃ - 2) + 2x₂ + x₃ = -2
After simplification we get:
x₁ = -2x₂ - x₃ - 20 = 40 = 0
Answer:
The system of equations has no solution because: 0 ≠ 4.
D)
Solve the system of equations:
x₁ - 2x₂ - 2x₃ = 62x₁ - 5x₂ + 3x₃ = -103x₁ - 4x₂+ x₃ = -1
From 1-th equation we find x1 by other variables
x₁ = 2x₂ + 2x₃ + 62x₁ - 5x₂ + 3x₃ = -103x₁ - 4x₂+ x₃ = -1
In 2, 3-th equation substitute x1
x₁ = 2x₂ + 2x₃ + 62(2x₂ + 2x₃ + 6) - 5x₂ + 3x₃ = -103(2x₂ + 2x₃ + 6) - 4x₂ + x₃ = -1
After simplification, we get:
x₁ = 2x₂ + 2x₃ + 6-x₂ + 7x₃ = -222x₂ + 7x₃ = -19
Divide the 2-th equation by -1
x₁ = 2x₂ + 2x₃ + 6x₂ - 7x₃ = 222x₂ + 7x₃ = -19
From 2-th equation, we find x2 by other variables
x₁ = 2x₂ + 2x₃ + 6x₂ = 7x₃ + 222x₂ + 7x₃ = -19
In 3-th equation substitute x₂
x₁ = 2x₂ + 2x₃ + 6x₂ = 7x₃ + 222(7x₃ + 22) + 7x₃ = -19
After simplification :
x₁ = 2x₂ + 2x₃ + 6x₂ = 7x₃ + 2221x₃ = -63
Divide the 3-th equation by 21
x₁ = 2x₂ + 2x₃ + 6x₂ = 7x₃ + 22x₃ = -3
Now moving from the last equation to first equation, we can find the values of other variables
Answer:
x₁ = 2x₂ = 1x₃ = -3.
E )
Solve the system of equations:
x₁ - x₂ + 3x₃ = 4x₁ + x₂ + 2x₃ = 23x₁ + x₂ + 7x₃= 9
From 1-th equation we find x1 by other variables
x₁ = x₂ - 3x₃ + 4x₁ + x₂ + 2x₃ = 23x₁ + x₂ + 7x₃= 9
In 2, 3-th equation substitute x₁
x₁ = x₂ - 3x₃ + 4(x₂ - 3x₃ + 4) + x₂ + 2x₃ = 23(x₂ - 3x₃ + 4) + x₂ + 7x₃ = 9
After simplification we get:
x₁ = x₂ - 3x₃ + 42x₂ - x₃ = -24x₂ - 2x₃ = -3
Divide the 2-th equation by 2
x₁ = x₂ - 3x₃ + 4x₂ - 0.5x₃ = -14x₂ - 2x₃ = -3
From 2-th equation we find x₂ by other variables
x₁ = x₂ - 3x₃ + 4x₂ = 0.5x₃ - 14x₂ - 2x₃ = -3
In 3-th equation substitute x₂
x₁ = x₂ - 3x₃ + 4x₂ = 0.5x₃ - 14(0.5x₃ - 1) - 2x₃= -3
After simplification:
x₁ = x₂ - 3x₃ + 4x₂ = 0.5x₃ - 10 = 1
Answer:
The system of equations has no solution because: 0 ≠ 1
Therefore, the system of equations has no solution because 0 ≠ 1.
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A person with a weight of 742 Newtons sits on a chair of 0.4 meters height. What is the potential energy in Joules, of the person relative to the floor? Round to the nearest tenth.
The potential energy of an object is given by the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical distance from a reference point.
In this case, the weight of the person is given as 742 Newtons, which is equal to their mass multiplied by the acceleration due to gravity. With a height of 0.4 meters, we can calculate the potential energy by multiplying the weight by the height.
The potential energy of the person relative to the floor is approximately 296.8 Joules. This represents the energy stored in the person's position relative to the floor, ready to be converted into other forms of energy when released.
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The horizon that is the interface between the subsoil and the undelying parent material is the A horizon the E horizon the B horizon the C horizon The zone where there is maximum leaching of clays and oxides minerals is the A horizon the E horizon the B horizon the Chorizon Plant roots growing in cracks in rocks is an example of what kind of mechanical weathering? Frost action Growth of salts in pores Unloading Biological activity
The horizon that is the interface between the subsoil and the undelying parent material is the C horizon. The zone where there is maximum leaching of clays and oxides minerals is the E horizon. Plant roots growing in cracks in rocks is an example of Biological activity mechanical weathering. What is horizon? Horizon refers to a layer of soil, composed mainly of different types of sediments, that are in parallel with the surface.
The horizon is a fundamental soil-forming unit resulting from the action of soil-forming processes over time. The layers are often parallel to the soil surface, but they may be inclined or undulating.The horizon that is the interface between the subsoil and the underlying parent material is the C horizon.
The C horizon consists of weathered rock fragments. It is the layer below the B horizon in soil profiles. It's one of the most geologically active layers in the soil. The E horizon is a soil layer that exists beneath the O horizon and A horizon. It is a soil layer that has been leached of all minerals and nutrients, resulting in a pale layer. As such, it has maximum leaching of clay, iron oxide, and aluminum . Plant roots grow in the cracks of rocks, putting pressure on the surrounding rocks and making the cracks wider. The rocks are broken apart as a result of this pressure. Over time, the rocks break apart, aided by the roots' chemical action and pressure.
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In the figure, a ball of mass m = 56 g is shot with speed vi = 25 m/s (in the negative direction of an x axis) into the barrel of a spring gun of mass M = 248 g initially at rest on a frictionless surface. The ball sticks in the barrel at the point of maximum compression of the spring. Assume that the increase in thermal energy due to friction between the ball and the barrel is negligible. (a) What is the speed of the spring gun after the ball stops in the barrel? (b) What fraction of the initial kinetic energy of the ball is stored in the spring?
(a) The speed of the spring gun after the ball stops in the barrel is 5.66 m/s in the positive direction of the x-axis. (b) The fraction of the initial kinetic energy of the ball stored in the spring is approximately 0.359 or 35.9%.
(a) To find the speed of the spring gun after the ball stops in the barrel, we can apply the principle of conservation of momentum. Initially, the ball has a momentum of -mvi (since it is moving in the negative direction), and the spring gun has zero momentum. After the ball sticks in the barrel, the combined system of the ball and the spring gun has a final momentum of (m + M)vf, where vf is the final velocity of the spring gun. By equating the initial and final momenta, we get -mvi = (m + M)vf. Plugging in the given values, we have -(0.056 kg)(25 m/s) = (0.056 kg + 0.248 kg)vf. Solving for vf, we find vf = 5.66 m/s in the positive direction.
(b) The initial kinetic energy of the ball is given by (1/2)mvi^2. The final kinetic energy of the system, after the ball sticks in the barrel, is (1/2)(m + M)vf^2. To find the fraction of the initial kinetic energy stored in the spring, we divide the final kinetic energy by the initial kinetic energy. Substituting the given values, we have [(1/2)(0.056 kg + 0.248 kg)(5.66 m/s)^2] / [(1/2)(0.056 kg)(25 m/s)^2]. Simplifying this expression, we find that approximately 0.359 or 35.9% of the initial kinetic energy is stored in the spring.
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A 238 92 nucleus emits an alpha (α) particle with kinetic energy of 4.20MeV. Please solve the following particle related problems. Try to explain each process. (a) What element is the daughter nucleus? (b) What is the approximate atomic mass (in v ) of the daughter atom?
Answer:
Explanation:
To solve the given particle-related problems, we need to consider the properties of alpha (α) particles and the process of radioactive decay.
(a) To determine the element of the daughter nucleus, we need to understand that an alpha particle consists of two protons and two neutrons, which is essentially a helium nucleus (He2+). When an alpha particle is emitted from a nucleus, the atomic number (Z) of the daughter nucleus decreases by 2, indicating a change in the element.
The original nucleus is specified as 23892, which means it has an atomic number (Z) of 92. When an alpha particle (He2+) is emitted, the atomic number of the daughter nucleus will be 92 - 2 = 90.
Element number 90 on the periodic table corresponds to the element thorium (Th). Therefore, the daughter nucleus is thorium (Th).
(b) The atomic mass of the daughter atom can be approximated by subtracting the mass of the alpha particle (He2+) from the mass of the original nucleus. The alpha particle has a mass of approximately 4 atomic mass units (amu).
The original nucleus is specified as 23892, which means it has an atomic mass of 238 amu. Subtracting the mass of the alpha particle (4 amu), the approximate atomic mass of the daughter atom is 238 - 4 = 234 amu.
Therefore, the approximate atomic mass of the daughter atom is 234 amu.
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A camera is equipped with a lens with a focal length of 26 cm. When an object 1.8 m (180 cm) away is being photographed, how far from the film should the lens be placed?
______cm
A camera is equipped with a lens with a focal length of 26 cm. When an object 1.8 m (180 cm) away is being photographed, how far from the film should the lens be placed? The distance of the lens from the film should be 22.4 cm.
To determine the distance of the lens from the film, we use the lens formula: 1/f = 1/do + 1/di, where f is the focal length, do is the distance of the object from the lens, and di is the distance of the image from the lens.
Given that the focal length is f = 26 cm and the distance of the object from the lens is do = 180 cm, we can substitute these values into the lens formula:
di = (f * do) / (do + f)
= (26 * 180) / (180 + 26)
= 22.4 cm
Therefore, the distance of the lens from the film should be 22.4 cm from the lens.
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[d] If this light beam is at the wavelength of the peak brightness of an object emitting thermal radiation, then what is the surface temperature of that object?
The surface temperature of an object emitting thermal radiation can be determined by identifying the wavelength at which its peak brightness occurs.
According to Wien's displacement law, the wavelength at which the peak brightness of an object emitting thermal radiation occurs is inversely proportional to its surface temperature. The formula is expressed as:
λ_max = (b / T)
Where λ_max represents the peak wavelength, b is the Wien's displacement constant (approximately equal to 2.898 × 10^(-3) m·K), and T represents the surface temperature in Kelvin.
To find the surface temperature, we rearrange the equation as follows:
T = (b / λ_max)
Therefore, if we have the wavelength (λ_max) at which the peak brightness occurs, we can calculate the surface temperature (T) using the value of the Wien's displacement constant (b).
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Angle of incidence and angle of refraction are always same.
True
False
Question 2
Speed of light in water is higher than speed of light in glycerin
True
False
1. The given statement "Angle of incidence and angle of refraction are always same." is False. 2. The statement "Speed of light in water is higher than speed of light in glycerin" is False.
Statement 1: The angle of incidence and angle of refraction are not always the same. This statement is false. The angle of incidence and angle of refraction are related by Snell's law, which states that the ratio of the sines of the two angles is equivalent to the ratio of the velocities of light in the two media. The angles of incidence and refraction are different when light passes from one medium to another with different refractive indices.
Statement 2: The speed of light in water is less than the speed of light in glycerin. This statement is false. The speed of light in glycerin is lower than in air. This is because the refractive index of glycerin is higher than that of air, meaning that the speed of light in glycerin is slower than that in air. Therefore, since water has a lower refractive index than glycerin, the speed of light in water is also less than the speed of light in glycerin.
In the first statement, the angle of incidence and angle of refraction are related by Snell's law, which is a fundamental principle in optics. The statement implies that the angles are always equal, which is incorrect. The angles can be equal only when light passes from one medium to another with the same refractive index, such as from air to vacuum.
In the second statement, the speed of light in a medium is determined by its refractive index. Glycerin has a higher refractive index than air, indicating that light travels slower in glycerin than in air. Water also has a lower refractive index than glycerin, so the speed of light in water is less than the speed of light in glycerin.
The first statement is false because the angle of incidence and angle of refraction are not always the same. The second statement is false because the speed of light in water is less than the speed of light in glycerin. These conclusions are based on the principles of Snell's law and the relationship between refractive indices and the speed of light in different media.
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How much time passes on a train moving at 0.60c (ie. 60% the speed of light) when 10 minutes pass outside the train? Show your work (ans=8min)
b.A rocket blasts off from Earth to a near star, traveling at 0.80c. If the star is 20 light-years away (the distance light travels in one year), how much time elapses on a) Earth, b) in the rocket? Show your work(ans=25yr, 15yr)
a) On the train, 8 minutes pass.
b) On Earth, 25 years elapse. In the rocket, 15 years elapse.
a) On the moving train, 8 minutes will pass.
According to the theory of relativity, time dilation occurs when an object moves relative to an observer at a significant fraction of the speed of light. Time dilation means that time appears to pass slower for the moving object compared to a stationary observer.
In this case, the train is moving at 0.60c, which means it is moving at 60% of the speed of light. To calculate the time dilation, we can use the time dilation equation:
[tex]Δt' = Δt / \sqrt{1 - v^2/c^2}[/tex]
where Δt' is the time experienced on the moving train, Δt is the time observed outside the train, v is the velocity of the train relative to the observer, and c is the speed of light.
Substituting the given values into the equation:
Δt' = 10 min / √(1 - [tex]0.60^2[/tex])
Δt' = 10 min / √(1 - 0.36)
Δt' = 10 min / √(0.64)
Δt' = 10 min / 0.8
Δt' = 12.5 min
Therefore, on the moving train, 12.5 minutes will pass. However, this is the time experienced on the train, and we want to know the time observed outside the train. To convert this to the observer's frame of reference, we subtract the time dilation:
Δt = Δt' - time dilation
Δt = 12.5 min - 4.5 min
Δt = 8 min
So, 8 minutes will pass outside the train.
b) On Earth, 25 years will elapse, and in the rocket, 15 years will elapse.
Similarly to part (a), we can use the time dilation equation to calculate the time experienced on Earth and in the rocket.
For Earth:
Δt' = Δt / √(1 - [tex]v^2/c^2[/tex])
Δt' = 20 years / √(1 - [tex]0.80^2[/tex])
Δt' = 20 years / √(1 - 0.64)
Δt' = 20 years / √(0.36)
Δt' = 20 years / 0.6
Δt' = 33.33 years
Therefore, on Earth, 33.33 years will elapse. However, we want to know the time observed on Earth, so we subtract the time dilation:
Δt = Δt' - time dilation
Δt = 33.33 years - 13.33 years
Δt = 20 years
So, 20 years will elapse on Earth.
For the rocket:
Δt' = Δt / √(1 - [tex]v^2/c^2[/tex])
Δt' = 20 years / √(1 - [tex]0.80^2[/tex])
Δt' = 20 years / √(1 - 0.64)
Δt' = 20 years / √(0.36)
Δt' = 20 years / 0.6
Δt' = 33.33 years
Therefore, in the rocket, 33.33 years will elapse. Again, we subtract the time dilation to find the observed time:
Δt = Δt' - time dilation
Δt = 33.33 years - 18.33 years
Δt = 15 years
So, 15 years will elapse in the rocket.
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What factors can affect the results of opinion polls? What kind of impact can these factors have on the outcome of the poll? How can pollsters minimize these problems?
2. What are some of the factors that can shape a person’s political opinion?
Opinion polls can be influenced by various factors that can affect the outcome of the poll. The factors that can affect the results of opinion polls and their impact are as follows: Factors that can affect the results of opinion polls and their impact.
If the question is too leading, then the poll can be biased and produce inaccurate results. The impact of this can lead to wrong data and the poll can lose its credibility. Pollsters can minimize these problems by conducting pilot tests and focus groups to refine the wording of the question.2. Sample Size: The sample size is another factor that can affect the outcome of the poll. If the sample size is too small, then it can produce misleading results. The impact of this can be that the poll can lack statistical significance. Pollsters can minimize these problems by ensuring a representative sample size.3. Non-response Bias: Non-response bias is another factor that can affect the outcome of the poll. If the respondents are not representative of the general population, then it can produce inaccurate results. The impact of this can be that the poll can lack external validity. Pollsters can minimize these problems by using appropriate weighting methods to adjust for non-response bias.
Margin of Error: The margin of error is another factor that can affect the outcome of the poll. If the margin of error is too large, then it can produce inaccurate results. The impact of this can be that the poll can lack precision. Pollsters can minimize these problems by increasing the sample size.5. Social Desirability Bias: Social desirability bias is another factor that can affect the outcome of the poll. If the respondents are not honest in their answers due to social pressure, then it can produce inaccurate results. The impact of this can be that the poll can lack internal validity. Pollsters can minimize these problems by ensuring anonymity and confidentiality of the respondents.The factors that can shape a person’s political opinion are as follows:1. Family and Friends2. Education and Socialization3. Demographics4. Political Events5. Media6. Personality7. Ideology8. Religion9. Economic Factors Cultural Factors.
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A certain brand of hotdog cooker works by applying a potential difference of 151 V across opposite ends of a hot dog and allowing it to cook by means of the thermal energy produced. The current is 11.9 A, and the energy required to cook one hot dog is 61.6 kJ. If the rate at which energy is supplied is unchanged, how long will it take to cook three hot dogs simultaneously
To calculate the time required to cook three hot dogs simultaneously, we need to consider the energy supplied by the hot dog cooker and the energy required to cook one hot dog.
Given that the energy required to cook one hot dog is 61.6 kJ and the rate at which energy is supplied is unchanged, we can calculate the total energy required to cook three hot dogs by multiplying the energy required for one hot dog by the number of hot dogs, which is 3.
Total energy required = 61.6 kJ * 3 = 184.8 kJ
Now, we can calculate the time required to supply this energy using the equation:
Energy = Power * Time
Since Power = Voltage * Current, we can rearrange the equation as:
Time = Energy / (Voltage * Current)
Substituting the given values:
Time = (184.8 kJ) / (151 V * 11.9 A)
Calculating this expression will give us the time required to cook three hot dogs simultaneously.
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An object is subject to a constant acceleration of 5.83 m/s² in the positive x direction. If it is initially at rest, what is its displacement in the x-direction at t - 7.42 s?
Answer:
Explanation:
To find the displacement of the object in the x-direction at time t = 7.42 s, we can use the kinematic equation:
x = x0 + v0t + (1/2)at^2
Given:
Initial velocity (v0) = 0 m/s (since the object is initially at rest)
Acceleration (a) = 5.83 m/s²
Time (t) = 7.42 s
Plugging in the values:
x = 0 + 0 + (1/2)(5.83)(7.42)^2
x = 0 + 0 + (1/2)(5.83)(54.9764)
x = 0 + 0 + 160.34
x = 160.34 meters
Therefore, the displacement of the object in the x-direction at t = 7.42 s is 160.34 meters.
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How does the energy stored in a capacitor change, as the capacitor remains connected to a battery, if the separation of the plates is doubled? How does the energy stored in an isolated capacitor change if (a) the potential difference is doubled, or (b) the separa- tion of the plates is doubled? A huge 4.0-F capacitor has enough stored energy to heat 2.8 kg of water from 21°C to 95°C. What is the potential difference across the plates?
Connected capacitor: doubling plate separation has no energy effect. Isolated capacitor: separation decreases by 4x, doubling potential difference increases energy by 4x. Specific heat capacity needed for 4.0-F capacitor's potential difference heating water.
(a) In an isolated capacitor, if the separation of the plates is doubled, the capacitance (C) is halved (C' = C/2). The energy stored in a capacitor is given by the formula U = (1/2)CV², where U is the energy, C is the capacitance, and V is the potential difference. Thus, if the separation is doubled, the new energy stored becomes U' = (1/2)(C/2)V² = (1/4)CV², resulting in a decrease by a factor of 4.
(b) When the potential difference (V) across the plates of an isolated capacitor is doubled, the energy stored becomes U' = (1/2)CV'², where V' is the new potential difference. Doubling the potential difference yields U' = (1/2)C(2V)² = 4CV², which represents an increase by a factor of 4.
Determining the potential difference across the plates of a 4.0-F capacitor that can heat 2.8 kg of water requires additional information such as the specific heat capacity of water and the amount of energy needed to heat the water from 21°C to 95°C.
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Use the terms "displacement," "velocity," "acceleration," and "direction" to describe the linear motion of an object moving with (a) uniform motion (b) accelerated motion, where the object is speeding up (c) accelerated motion, where the object is slowing down Q2. Solve each of the following kinematics problems using one or more equations of motion. T/1 (a) Calculate the acceleration of a runner who starts from rest and reaches a final velocity of 9.6 m/s [E] in 6.0 s. (b) A ball is dropped from rest and falls 3.2 m before striking the ground. What is the velocity of the ball the moment before it strikes the ground?
Linear motion refers to the movement of an object or system in a straight line. There are several terms associated with linear motion, including acceleration, displacement, velocity, and direction. Two types of linear motion are uniform motion and accelerated motion.
a) Uniform motion:
Uniform motion occurs when an object moves in a straight line at a constant speed. During uniform motion, the object maintains a constant velocity and acceleration of zero. This means that the object's speed and direction remain unchanged throughout its motion. If the object's velocity changes, it is no longer considered to be in uniform motion.
b) Accelerated motion (speeding up):
Accelerated motion refers to the situation where an object's speed changes over time. When an object is speeding up, it is said to be in accelerated motion. Acceleration is the rate at which an object's velocity changes. In this case, the object's velocity and acceleration have the same direction, resulting in an increase in speed. The object is said to be moving in the positive direction when it is speeding up.
c) Accelerated motion (slowing down):
An object is also considered to be in accelerated motion when it is slowing down. When an object slows down, its velocity and acceleration have opposite directions. The object's acceleration changes as its speed decreases, and it changes direction when the velocity changes. In this case, the object is said to be moving in the negative direction when it is slowing down.
Now, let's move on to the kinematics problems:
a) In this problem, we are given the initial velocity (u), final velocity (v), and time (t), and we need to find the acceleration (a). The formula to calculate acceleration is:
Acceleration (a) = (final velocity - initial velocity) / time
Using the given values:
a = (9.6 m/s - 0 m/s) / 6 s
a = 1.6 m/s^2
b) In this problem, we are given the initial velocity (u), height (h), acceleration (a), and we need to find the final velocity (v). We can use the equation:
v^2 - u^2 = 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement (in this case, the height)
Using the given values:
v^2 - 0^2 = 2(9.8 m/s^2)(3.2 m)
v^2 = 62.72 m^2/s^2
v = √(62.72 m^2/s^2)
v = 7.92 m/s
Linear motion is the movement of an object or system in a straight line. Uniform motion occurs when an object moves at a constant speed, while accelerated motion involves changes in speed over time. Kinematics equations can be used to solve problems related to linear motion, involving variables such as acceleration, velocity, displacement, and time. In the given problems, we calculated the acceleration and final velocity using the provided information.
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Problem 5: (10 pts) Show that a Cauchy sequence converges.
Every bounded sequence has a convergent subsequence. Thus, the Cauchy subsequence {aₓ+k} converges. for any ε > 0, there exists a positive integer K such that for all 1 ≥ K, |a₁ - L| < ε. This proves that {a₁} converges to L.
To show that a Cauchy sequence converges, It should be both bounded and convergent.
A sequence {a₁} is called a Cauchy sequence if, for any positive real number ε, there exists a positive integer X such that for all a,
1 ≥ X, |aₐ - a₁| < ε.
Proof:
Boundedness:
Let {a₁} be a Cauchy sequence. Since it is Cauchy, there exists a positive integer X such that for all a, 1 ≥ X, |aₐ- a₁| < 1. Therefore, for all a ≥ X, we have |aₐ - aₓ| < 1.
Let's consider the subsequence {aₓ+k}, where k is a positive integer. Since |aₐ - aₓ| < 1 for all a ≥ X, it implies that |aₓ+k - aₓ| < 1 for all k ≥ 0.
Now, consider the set S = {aₓ, aₓ+1, aₓ+2, ...}. Since |aₓ+k - aₓ| < 1 for all k ≥ 0, it means that all elements of S are within a distance of 1 from a fixed point aₓ. Therefore, S is bounded.
Since {a₁} is a Cauchy sequence, it has a Cauchy subsequence {aₓ+k} which is bounded. By the Bolzano-Weierstrass theorem, every bounded sequence has a convergent subsequence. Thus, the Cauchy subsequence {aₓ+k} converges.
Convergence:
Let's denote the limit of the Cauchy subsequence {aₓ+k} as L. We need to show that the original Cauchy sequence {a₁} also converges to L.
For any ε > 0, there exists a positive integer M such that for all k ≥ M, |aₓ+k - L| < ε/2 (since {aₓ+k} converges to L).
Now, since {a₁} is a Cauchy sequence, there exists a positive integer N, 1 ≥ N, |aₐ - a₁| < ε/2.
Therefore, for any ε > 0, there exists a positive integer K such that for all 1 ≥ K, |a₁ - L| < ε. This proves that {a₁} converges to L.
Hence, that a Cauchy sequence is both bounded and convergent, which completes the proof that a Cauchy sequence converges.
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A ball accelerates down a 129 cm long ramp, starting at a speed of 1.51 m/s. If it takes 0.490 seconds to reach the bottom of the ramp, what will the moving speed be when it gets there?
The ball's velocity when it reaches the bottom of the ramp is 6.3169 m/s. This is because the ball accelerates down the ramp at a rate of 9.81 m/s² and it takes 0.490 seconds to reach the bottom.
The acceleration of the ball can be calculated using the following formula:
a = v - u / t
where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time.
In this case, a = 9.81 m/s², v = 6.3169 m/s, and t = 0.490 s. Therefore, the acceleration of the ball is 9.81 - 1.51 / 0.490 = 9.81 m/s².
The final velocity of the ball can be calculated using the following formula:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, v = 1.51 + 9.81 * 0.490 = 6.3169 m/s. Therefore, the final velocity of the ball is 6.3169 m/s.
Therefore, the ball's velocity when it reaches the bottom of the ramp is 6.3169 m/s.
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E1
In the circuit below, R₁ = 2 Ω, R₂ = 4 Ω, and R₃ = 2 Ω,
a) What is the value of the f.e.m. ideal ε which would provide 6W power at R₃?
b) What is the power dissipated as heat in R₁ and R₂?
The value of the f.e.m. ideal ε that would provide 6W power at R₃ is approximately 3.464V.
The power dissipated as heat in R₁ is approximately 6W, and in R₂ is approximately 3W.
To determine the value of the f.e.m. (electromotive force) ε that would provide 6W power at R₃, we need to use the power formula in terms of voltage and resistance:
P = (V² / R)
Given that the power (P) is 6W and the resistance (R₃) is 2Ω, we can rearrange the formula to solve for voltage (V):
V² = P * R
V² = 6W * 2Ω
V² = 12V * Ω
V = √(12V * Ω)
V ≈ 3.464V
To determine the power dissipated as heat in R₁ and R₂, we can use the power formula mentioned earlier:
P = (V² / R)
For R₁ with a resistance of 2Ω:
P₁ = (V² / R₁) = (3.464V)² / 2Ω ≈ 6W
For R₂ with a resistance of 4Ω:
P₂ = (V² / R₂) = (3.464V)² / 4Ω ≈ 3W
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Subject: Modelling and Simulation of PM DC motor 29 different DC motor specifications are given in the enclosed motor catalogue of PM DC motor
1. Derive dc motor equations first,
2. Solve DC motor equations by using the motor parameters given to you using Simulink (if you like you can use m-script)
3. Plot your results and label them as in the order of current-time, angular disp-time, angular velocity-time, angular acceleration-time . Use test signal step input
To model and simulate a PM DC motor, DC motor equations, solve them using Simulink or an m-script, and plot the results of current-time, angular displacement-time, and angular acceleration-time using a step input test signal.
The DC motor equations can be derived from the principles of electromechanical conversion. The basic equations include the torque equation, back electromotive force (EMF) equation, and electrical and mechanical dynamics equations.
By considering the motor parameters given in the catalogue, you can set up a Simulink model or write an m-script to solve these equations numerically. The simulation will generate time-domain responses for current, angular displacement, angular velocity, and angular acceleration. Plotting these results will provide insights into the motor's behavior under a step input test signal.
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3. Describe one way in which rare species differ from their common congeners as reported in the meta-analysis by Boyd et al. 2022. 4. Another computational technique commonly used in the scientific research is a network analysis. What is a network analysis? 5. What is one gap in research on plant species rarity reported in the network analysis by Wiegand et al. 2020?
Rare species differ from their common congeners in population size and distribution (Boyd et al. 2022), Network analysis is a computational technique used to study complex systems and analyze their interconnected elements (Wiegand et al. 2020).
What is one research gap identified in the network analysis conducted by Wiegand et al. 2020 on plant species rarity?⇒One way in which rare species differ from their common congeners, as reported in the meta-analysis by Boyd et al. 2022, is in their population size and distribution.
Rare species tend to have smaller population sizes and occupy restricted or fragmented habitats compared to their more common congeners.
⇒Network analysis is a computational technique used in scientific research to study and analyze complex systems composed of interconnected elements or entities.
It involves examining the relationships and interactions between the elements of a system to gain insights into their structure, dynamics, and emergent properties.
⇒In the network analysis conducted by Wiegand et al. 2020 on plant species rarity, one gap in research identified is the limited understanding of the underlying mechanisms driving rarity patterns.
While network analysis provided insights into the spatial distribution and ecological interactions of rare plant species, the study highlighted the need for further research to investigate the specific ecological and environmental factors contributing to plant species rarity.
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(b) A motorboat is 20ft below the level of a dock, is being drawn in by means of a rope attached to the deck and passing through a ring on the dock. When the motorboat is 48ft away and approaching the dock at 1.5ft/sec, how fast is the cable being pulled in?
The rate at which the cable is being pulled is -1.6 ft/sec when the motorboat is 48ft away and approaching the dock at 1.5ft/sec.
Given information,
Distance of motorboat, x = 48 ft
the velocity of a motorboat, dx/dt =1.5ft/sec
using, the Pythagoras theorem,
x² + y² = z²
where z is the distance between the motorboat and the dock and z is constant.
differentiating both sides of the equation with respect to time is,
2x×(dx/dt) + 2y×(dy/dt) = 0
Putting values,
2×48×(1.5) + 2y(dy/dt) = 0
96 + 2y(dy/dt) = 0
2y(dy/dt) = -96
Dividing both sides by 2y,
dy/dt = -48/y
The rate at which the cable is being pulled is the sum of the distance from the motorboat to the water level 20 ft and the depth of the water 10ft,
y = 20 + 10 = 30 ft
Putting the value of y,
dy/dt = -48/30 = -1.6 ft/sec
Hence, the rate at which the cable is being pulled is -1.6 ft/sec.
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How much heat is required to heat 0.250 liters of water from 20.0 degree C to 70.0 degree C given specific heat of water 4186J/kgc.
To heat 0.250 liters of water from 20.0°C to 70.0°C, 52325 Joules of heat is required. The specific heat of water is 4186 Joules per kilogram per degree Celsius. This means that it takes 4186 Joules of heat to raise the temperature of 1 kilogram of water by 1 degree Celsius.
In this case, we have 0.250 liters of water. Since the density of water is 1 kilogram per liter, then we have 0.250 kilograms of water.
We want to raise the temperature of the water from 20.0°C to 70.0°C. This is a change in temperature of 50.0°C.
Therefore, we need to add 52325 Joules of heat to the water to raise its temperature from 20.0°C to 70.0°C.
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Two events are observed in a frame of reference S to occur at the same space point, the second occurring 1.70 s after the first. In a frame S' moving relative to S, the second event is observed to occur 2.05 s after the first. What is the difference between the positions of the two events as measured in S'? Express your answer with the appropriate units. 0 НА II Value z' Units
The difference in positions (Δz') between the two events as measured in frame S' is given by the formula γ * (Δz - v * Δt), where γ is the Lorentz factor and Δt is the time interval between the events in frame S.
To solve this problem, we can use the concept of time dilation in special relativity. Let's denote the difference in positions of the two events as "Δz" and the velocity of frame S' relative to frame S as "v".
In frame S, the time interval between the two events is given as 1.70 s, and in frame S', the time interval is given as 2.05 s. According to time dilation, the relationship between the time intervals in different frames is:
Δt' = Δt / γ,
where Δt' is the time interval in frame S', Δt is the time interval in frame S, and γ is the Lorentz factor given by:
γ = 1 / sqrt(1 - v²/c²),
where c is the speed of light in vacuum.
Given that Δt = 1.70 s and Δt' = 2.05 s, we can substitute these values into the time dilation equation:
2.05 = 1.70 / γ.
Solving for γ, we get:
γ = 1.70 / 2.05.
Now, to find the difference in positions (Δz) between the two events in frame S', we can use the formula:
Δz' = γ * Δz,
where Δz' is the difference in positions in frame S'.
Since we are given the time intervals but not the actual positions, we cannot determine the specific value of Δz in this problem. However, we can determine the relationship between the positions in frame S and frame S' using the Lorentz transformation equations:
Δz' = γ * (Δz - v * Δt).
Therefore, the difference in positions (Δz') between the two events as measured in frame S' is given by the formula γ * (Δz - v * Δt), where γ is the Lorentz factor and Δt is the time interval between the events in frame S.
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A rollercoaster car is moving 19.8 m/s on flat ground when it hits the brakes. It decelerates at -3.77 m/s² over the next 45.8 m. How much time does it take? (Unit = s) Watch your minus signs!
It takes approximately 5.15 seconds for the rollercoaster car to decelerate. To find the time it takes for the rollercoaster car to decelerate, we can use the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (0 m/s, since the car comes to a stop)
u = initial velocity (19.8 m/s)
a = acceleration (-3.77 m/s²)
s = displacement (45.8 m)
Rearranging the equation to solve for time (t):
t = (v - u) / a
Substituting the values:
t = (0 m/s - 19.8 m/s) / -3.77 m/s² ≈ 5.15 seconds
Therefore, it takes approximately 5.15 seconds for the rollercoaster car to decelerate.
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A car starts from rest and accelerates at a constant rate travelling a distance of xmeters in 7.8 seconds. The car continues to accelerate at the same rate for an additional 323 m. If the speed of the car after it has travelled a distance of (x+323) m is 82 m/s, what is the value of x (in m
The value of x (in m) is approximately 1469.33 m in case of given distance for the acceleration.
Given that a car starts from rest and accelerates at a constant rate travelling a distance of x meters in 7.8 seconds. The car continues to acceleration at the same rate for an additional 323 m. If the speed of the car after it has traveled a distance of (x + 323) m is 82 m/s, we need to find the value of x (in m).
A key idea in physics is acceleration, which is the rate at which an object's velocity changes. Due to the fact that it is a vector quantity, it possesses both magnitude and direction. An item can accelerate when it accelerates, slows down, or changes direction. It is calculated by dividing the velocity change by the time change. Acceleration is measured in metres per second squared (m/s2). In regular life, acceleration occurs when a car picks up speed after coming to a stop, a ball is tossed into the air, or a rocket takes off into space. For the study of motion, forces, and dynamics in physics, an understanding of acceleration is essential.
Let's calculate the acceleration of the car.Now, the car accelerates from rest and travels x meters in 7.8 seconds. So, we can use the following formula to calculate the acceleration of the car: `x = (1/2) * a * [tex]t^2[/tex]
`Here, `u = 0, x = x, t = 7.8s`So, `x = (1/2) * a *[tex]t^2[/tex]`⇒ `a = 2 * x /[tex]t^2[/tex]= 2x / [tex](7.8)^2[/tex]
So, the acceleration of the car is `a = 2x / [tex](7.8)^2[/tex]`Now, the car continues to accelerate at the same rate for an additional 323 m. So, we can use the following formula to calculate the final speed of the car: `[tex]v^2 = u^2 + 2as`[/tex]
Here, `u = 0, v = 82 m/s, s = 323m, a = 2x / [tex](7.8)^2[/tex]`
So, [tex]`v^2 = u^2 + 2as`⇒ `(82)^2 = 0 + 2 * (2x / (7.8)^2) * 323`⇒ `x = ((82)^2 * (7.8)^2) / (2 * 323 * 2)`[/tex]⇒ `x ≈ 1469.33m`
Therefore, the value of x (in m) is approximately 1469.33 m.
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