Answer:
The question of analog vs. digital audio is one of the more hotly debated questions in the world of music, film and media today. Does digital sound better? Does analog sound better? Is there even a noticeable difference?
It’s impossible to understand the difference completely without understanding what distinguishes analog audio from digital audio. A full discussion of these terms is best left to your curriculum and discussions with your mentor in the studio. For now, though, here’s a brief explanation of what these two words mean, and the differences between them.
Analog refers to a continuously changing representation of a continuously variable quantity. Digital, however, refers to representing these variable quantities in terms of actual numbers, or digits. The last two sentences seem a bit complex, but let’s try to simplify them with an example. If you consider the numbers 1 and 2 on a number line, there are actually an infinite number of points between 1 and 2. This is what analog represents—the infinite number of possibilities between 1 and 2. Digital, on the other hand, only looks at certain number of fixed points along the line between 1 and 2 (for example, 1 ¼, 1 ½, 1 ¾, and 2).
Can you see the difference? Digital takes a few “snapshots” of the number line, while analog takes the whole line into account.
As another example, think of analog vs. digital as the difference between seeing something in real life and watching it on film. When we see something happen in real life, there are no “spaces” between what we see, so we’re watching it happen in analog. Film, however, is actually a series of still photographs that are taken in rapid-fire intervals, and when we see them in succession, it tricks our minds into thinking we’re seeing a continuous flow of movement. So in a manner of speaking, when we watch the event happen on film, we’re watching it digitally, because we’re watching increments of the event, rather than the whole thing in fluid motion. (Not to be confused with digital video vs. film, which is another discussion completely!)
Let’s bring this idea into audio, music, and the studio. Sound occurs naturally in analog–that is to say, sound occurs in a continuous set of waves that we hear with the human ear. (Think of it as a “wavy” line with an infinite number of points along it.) When we capture that sound in a way that represents all the possible frequencies, we’re recording in analog; when we use computers to translate the sound into a series of numbers that approximate what we’re hearing, we’re recording in digital.
Thus, a purely analog recording would be something that was recorded on tape and produced using manual equipment to mix, master and press into a vinyl LP. A purely digital recording would be recorded on a computer program such as Pro Tools, mixed, mastered and produced digitally, and eventually burned onto a CD as an MP3 or audio file.
The most ironic aspect of the debate about digital vs. analog recording is that nowadays a lot of music is a combination of the two. For example, you might record a song onto analog tape, but mix and master it digitally, or release it on the Internet as an MP3.
So what’s the difference in quality between analog and digital? The idea between digital recording is that our ears and brains technically can’t determine the spaces between the digital values, just like our brains interpret film as continuous motion. However, to many people, analog sound tends to be warmer, has more texture and is thought to capture a truer representation of the actual sound. Digital is felt to be somewhat cold, technical and perhaps lacking in analog’s nuance.
However digital is much cheaper. Recording an album with analog technology can require a whole studio full of equipment, but with digital recording technology, it’s possible to record a whole album in a bedroom on a laptop. And whereas analog technology can wear out or be damaged, digital media can last for an indefinite length of time.
Today many recording artists, both major and independent, record using a mixture of digital and analog techniques. While analog audio does give warmth and a truer sound quality, digital is cheaper to work with and offers more control over the finished product.
S1-CA-PHY41
The graph below represents the motion of a car during a 3.0 s time interval. What is the total displacement
of the object at the end of 3.0s
Answer:
The total displacement of the object is 20 m
Explanation:
Velocity vs Time Graph
It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.
The distance covered by the car can be calculated as the area behind the graph.
The total area can be calculated in two parts: One triangle from t=0 sec to t=2 sec, and one rectangle from t=2 sec to t=3 sec.
The area of a triangle is
A1= B*H/2, where B is the length of the base, and H is the length of the Height. The base is B=2 seconds, and the height is H = 10 m/s, thus:
A1= (2 sec)*(10 m/sec)/2=10 m
The area of a rectangle is
A2=W*L, where W is the width, L is the length. W=1 sec, L=10 m/s, thus:
A2=(1 sec)*(10 m/sec) = 10 m
Total displacement= 10 m + 10 m = 20 m
The total displacement of the object is 20 m
At what point of thermometric scale does Kelvin scale reading coincide with Fahrenheit scale
Answer:
At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.
Explanation:
We first need to find a relation between the Kelvin scale and the Fahranheit scale. We'll use the Celsius scale to relate them.
The Kelvin and Celsius scales are related by the formula:
K = °C + 273.15
Solving for °C:
°C = K - 273.15
Besides, the Kelvin and Celsius scales are related by:
°C = 5 ⁄ 9(°F - 32)
Now we find a temperature, say X, where both scales coincide. Equating both formulas:
X - 273.15=5 ⁄ 9(X - 32)
Multiply by 9:
9X - 2,458.35 = 5X - 160
Simplifying:
4X = 2,458.35 - 160=2,298.35
Solving:
X =2,298.35 / 4 = 574.59
At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.
What forces cause acceleration?
Answer:
friction, should be correct i think!
[04.05] An atom has an atomic number of 12 and a mass number of 25. How many neutrons does the atom have?
Answer: is 13
Explanation:
What are 3 parts of daltons atomic theory ?
A 40kg box sits on a shelf 0.5m above the floor. An employee moves the box to a height of 1.8m above the floor. How much work did the employee do in moving the box? A)20J B)72J C)200J D)520J E)720J
Answer:
b)72
Explanation
work done= force* distance
work done=40*1.8
72J
what would you want the after life to be like. examples are heaven and hell ,reincarnation ,eternal darkness , reliving your live but different choices or become a ghost,
Answer:
i would want to be a dog or a cat
Explanation:
there just funny
Which describes elements in the first group of the periodic table?
Answer:
Alkali metals are the chemical elements found in Group 1 of the periodic table. The alkali metals include: lithium, sodium, potassium, rubidium, cesium, and francium.
Explanation:
Answer:
alkali metals
Explanation:
Who do you think took Ms. C's lunch?
Answer:
the main character
Explanation:
i think
If the number of wavelengths that pass a point in a given time increases, what has increased?
A.frequency
B.amplitude
C.intensity
D.wavelength
A vector has an x-component
of 6.15 m and a y-component
of -3.88 m.
Find the direction of the vector.
8) Find the X and Y component of 10degree vector that has 5N.
Answer:
Fx = 4.92 [N]
Fy = 0.868 [N]
Explanation:
Let's take the 10 degrees as a measure from the horizontal component to the vector.
Thus taking the components in the X & y axes respectively:
Fx = 5*cos(10) = 4.92 [N]
Fy = 5*sin(10) = 0.868 [N]
what is science and examples with it
Answer:
science is the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.
Explanation:
A physics student drops a water balloon from an upstairs window. The window is 26.45 m
from the ground. How long does it take for the balloon to hit the ground? (Round to the
nearest tenth)
Answer:
Explanation:
Given:
Distance S = 26.45m
Using the equation of motion
S = ut+1/2gt²
26.45 = 0+1/2(9.81)t²
26.45 = 4.905t²
t² = 26.45/4.905
t² = 5.392
t = 2.528s
Hence the it took the balloon 2.52s to reach the ground
a blwoing ball traveling 2.0 m/s has a 16 J of Kinetic energy what is the mass of the bowling ball in KG
Answer:
ke=1/2mv2Explanation:
16=m(2)*2
16=m4
m=16÷4
m=4
where is the image located when an objectis placed 30cm from a convex lens with a focal length of 10cm?
Answer:
15 cm
Explanation:
using the thin lens formula for convex lens
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\\ \\where f = focal length \\ p = object distance \\ q = image distance\\[/tex]
so,
[tex]\frac{1}{q} = \frac{1}{f} - \frac{1}{p} \\\\\frac{1}{q} = \frac{1}{10} - \frac{1}{30} \\\frac{1}{q} = \frac{2}{30}\\\frac{1}{q} = \frac{1}{15cm}[/tex]
so,
q = 15 cm
What is the difference between a fact witness and an expert witness?
Answer:
Unlike a fact witness, an expert is entitled to compensation for participation in the case. A key distinction between a fact witness and an expert witness is that an expert witness may provide an opinion. Fact witnesses must limit their testimony to facts in regard to evidence they may have observed or been involved.
Explanation:
there
Answer:
Unlike a fact witness, an expert is entitled to compensation for participation in the case. A key distinction between a fact witness and an expert witness is that an expert witness may provide an opinion. Fact witnesses must limit their testimony to facts in regard to evidence they may have observed or been involved.
Explanation:
E2020
please awnser and show the ways
Answer:
Answers in solutions.
Explanation:
Question 6:
The density of gold is 19.3 g/cm³
The density of silver is 10.5 g/cm³
The density of the substance in Crown A;Density = mass ÷ volume = [tex]\frac{1930}{100}[/tex] = 19.3 g/cm³
Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.
The density of the substance in Crown B;Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043 g/cm³ ≈ 10.5 g/cm³ (answer rounded up to one decimal place)
Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.
The density of substance in Crown C;Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)
The density of the mixture:For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g
∴ for 1 cm³ of the mixture, its mass equal to [tex]\frac{29.8}{2}[/tex] = 14.9 g
Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.
* Crown C is not made up of a mixture of gold and silver.
Question 7:
An empty masuring cylinder has a mass of 500 g.Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.The total mass is now 850 gThe mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g
Density of the liquid (water) poured into the container = mass ÷ volume = 350 g ÷ 100 cm³ = 3.5g/cm³
Question 8:
A tank filled with water has a volume of 0.02 m³
(a) 1 liter = 0.001 m³
How many liters? = 0.02 m³ ?
Cross multiplying gives:
[tex]\frac{0.02 * 1}{0.001}[/tex] = 20 liters
(b) 1 m³ = 1,000,000 cm³
0.02 m³ = how many cm³ ?
Cross-multiplying gives;
[tex]\frac{0.02 * 1,000,000}{1}[/tex] = 20,000 cm³
(c) 1 cm³ = 1 ml
∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml
Question 9:
Caliper (a) measurement = 3.2 cm
Caliper (b) measurement = 3 cm
Question 10:
A stone is gently and completely immersed in a liquid of density 1.0 g/cm³in a displacement canThe mass of liquid which overflow is 20 gThe mass of the liquid which overflow = mass of the stone = 20 g
1 gram of the liquid occupies 1 cm³ of space.
20 g of the liquid will occupy; [tex]\frac{20 * 1}{1}[/tex] = 20 cm³
(a) Since the volume of the water displaced is equal to the volume of the stone.
∴ The volume of the stone = 20 cm³
(b) Mass = density × volume
Density of the stone = 5.0 g/cm³
Volume of the stone = 20 cm³
Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g
Explain the quantization of energy.
Answer:
The quantization of energy refers to the fact that at subatomic levels, energy is best thought of as occurring in discreet "packets" called photons. Like paper money, photons come in different denominations. ... Each photon contains a unique amount of discreet energy
List the 4 things technology can be.
Technology can be the four things listed below...
PhonesComputersTVsPlaystation or XboxAnswer:
mechanical
electronic
medical
industrail and manufacturing
Water is formed when two hydrogen atoms bond to an oxygen atom. The hydrogen and the oxygen in this example are different
They both are two different elements.SO the answer is Element
If a car that is moving 20.0 m/s has a momentum of 29000 kg·m/s, what mass is the car?
Answer:
Mass = 1450kg
Explanation:
P = M * V (where p is momentum, m is mass and v is velocity)
29000 = 20 * M
M = 29000 / 20
M= 1450 kg
When walking barefoot, Kevin can walk across the grass easily, but when he crosses the paved street, the street is hot and burns his feet. At the beach he walks quickly across the hot sand to the cool water. Why do some substances feel hot to his feet and others do not? Radiation is heat transfer through waves. Specific heat is the amount of energy required to raise the temperature of a substance 1 degree C.
You are designing an apparatus to shoot a water balloon through a 3rd story window (25m above the ground)
Answer:
The balloon needs a vertical velocity of approximately 22.14 m/s to reach the window
Explanation:
The given parameters are;
The destination of the designed water balloon we shoot = The 3rd story window
The height of the 3rd story window which the water balloon should reach = 25 m
Therefore, we have, the equation of motion of the water balloon given as follows;
v² = u² - 2 × g × s
Where;
[tex]u_y[/tex] = The initial vertical velocity of the balloon
[tex]v_y[/tex] = The final vertical velocity of the balloon = 0 m/s
g = The acceleration due to gravity = 9.8 m/s
s = The height the balloon is intended reach at the final velocity becomes 0 m/s = The height of the 3rd story window
∴ s = 25 m
Substituting the values, we have;
0² = [tex]u_y[/tex]² - 2 × 9.8 × 25
[tex]u_y[/tex]² = 2 × 9.8 × 25 = 490
[tex]u_y[/tex] = √490 = 7·√10
The initial vertical velocity of the balloon = [tex]u_y[/tex] = 7·√10 m/s
Therefore, the balloon needs a vertical velocity of 7·√10 m/s which is approximately 22.14 m/s to reach the window.
Using the formula for kinetic energy of a moving particle k=12mv2, find the kinetic energy ka of particle a and the kinetic energy kb of particle
b. remember that both particles rotate about the y axis. express your answers in terms of m, ω, and r separated by a comma.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The kinetic energy for particle a is [tex]K_a = \frac{9}{2} mr^2 w^2[/tex]
The kinetic energy for particle b is [tex]K_b = m w^2r^2 [/tex]
Explanation:
From the question we are told that
The mass of particle a is [tex]m[/tex]
The mass of particle b is [tex]2 m[/tex]
Generally the kinetic energy for particle a is mathematically represented as
[tex]K_a = \frac{1}{2} m v^2[/tex]
Here [tex]v_a[/tex] is the linear velocity of particle a which is mathematically represented as
[tex]v_b = w * 3r[/tex]
So
[tex]K = \frac{1}{2} m {w * 3r}^2[/tex]
=> [tex]K_a = \frac{9}{2} mr^2 w^2[/tex]
Generally the kinetic energy for particle a is mathematically represented as
[tex]K_b = \frac{1}{2} 2m v_b^2[/tex]
Here [tex]v_b[/tex] is the linear velocity of particle a which is mathematically represented as
[tex]v_b = w * r[/tex]
So
[tex]K_b = m (wr)^2[/tex]
=> [tex]K_b = m w^2r^2[/tex]
The kinetic energy of the particle in terms of angular speed (ω), mass of the object (m) and the radius of the path is [tex]\frac{1}{2} m \omega ^2 r^2[/tex].
The given parameters;
kinetic energy of the particle, K.E = ¹/₂mv²The kinetic energy of the of the particle in terms of the angular velocity is calculated as follows;
[tex]K_b = \frac{1}{2} m (\omega r)^2\\\\K_b = \frac{1}{2} m \omega ^2 r^2[/tex]
where;
m is the mass of the objectω is the angular speed of the objectr is the radius of the circular pathThus, the kinetic energy of the particle in terms of angular speed (ω), mass of the object (m) and the radius of the path is [tex]\frac{1}{2} m \omega ^2 r^2[/tex].
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16 William is sliding down a plastic slide on a playground. After he slides down the
ground, he feels a small electric shock when he touches a piece of metal. What has been built up on his body
that causes the shock?
A atoms
B electrons
C. neutrons
D. protons
Answer:Electrones
Explanation:Because its building up static
Answer:
B) Electrons
Explanation:
When William slides down the slide, the rubbing action between William and the slide causes electrons to be transferred to William and stored as a negative charge.
When he touches a piece of metal, the negative charge (i.e the build up of excess electrons), is released (discharged) into the metal causing what we know as a static shock.
On an icy day, you worry about parking your car in your driveway, which has an incline of 12º. Your neighbor's driveway has an incline of 9.0º, and the driveway across the street is at 6.0º. The coefficient of static friction between tire rubber and ice is 0.15. Which driveway(s) will be safe to park in?
Answer:
Driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.
Explanation:
Let suppose that car is represented by a particle, then we proceed to construct its free body diagram and corresponding equations of equilibrium:
[tex]\Sigma F_{x'} = f - W\cdot \sin \theta = 0[/tex] (Eq. 1)
[tex]\Sigma F_{y'} = N-W\cdot \cos \theta = 0[/tex] (Eq. 2)
Where:
[tex]f[/tex] - Static friction force, measured in newtons.
[tex]N[/tex] - Normal force on the car from the ground, measured in newtons.
[tex]W[/tex] - Weight of the car, measured in newtons.
[tex]\theta[/tex] - Driveway inclination, measured in sexagesimal degrees.
By applying definitions of maximum static friction force and weight, we expand the system of equations presented above:
[tex]\mu_{s}\cdot N -m\cdot g\cdot \sin \theta = 0[/tex] (Eq. 1b)
[tex]N - m\cdot g \cdot \cos \theta = 0[/tex] (Eq. 2b)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]m[/tex] - Mass of the car, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
Then, we substitute normal force and simplify the resulting expression:
[tex]\mu_{s}\cdot m\cdot g \cdot \cos \theta -m\cdot g \cdot \sin \theta = 0[/tex]
[tex]\mu_{s} = \tan \theta[/tex]
[tex]\theta = \tan^{-1}\mu_{s}[/tex] (Eq. 3)
Based on such result, we can conclude that if driveway inclination is greater than value reported, then car shall not be safe in case of parking. Our reference angle is: ([tex]\mu_{s} = 0.15[/tex])
[tex]\theta = \tan^{-1} 0.15[/tex]
[tex]\theta \approx 8.531^{\circ}[/tex]
By direct comparison, we find that driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.
At a drag race, a jet car travels 1/4 mile in 5.2 seconds. What is the final speed of the
car and its acceleration?
Answer:
1609.3
Explanation:
An observer recorded the following data for the motion of a car undergoing constant acceleration.
Time (s)
3.0
5.0
6.0
Speed (m/s)
4.0
7.0
8.5
What was the magnitude of the acceleration of the car?
4.5 m/s2
O 2.0 m/s2
3.0 m/s2
1.5 m/s2
Option D) 1.5 m/s² is the correct answer.
The magnitude of the acceleration of the car is 1.5 m/s².
given the data in the question;
record of the observation;
Time(s) Speed(m/s)
3.0 4.0
5.0 7.0
6.0 8.5
Since its undergoing constant acceleration, we can make use any two from the three data recorded.
So, lets consider motion between t₁ = 3.0s and t₂ = 5.0s
Initial velocity at t₁ ; u = 4.0 m/s
final velocity at t₂; v = 7.0 m/s
Now, Time Elapsed; [tex]t = t_2 - t_1 = 7.0s - 5.0s = 2.0s[/tex]
To determine the magnitude of acceleration "a", we use the first equation of motion which says;
[tex]v = u + at[/tex]
we make "a" the subject of the formula
[tex]a =\frac{v - u}{t}[/tex]
so we substitute in our values;
[tex]a = \frac{(7.0m/s) - (4.0m/s)}{2.0s}[/tex]
[tex]a = \frac{3.0m/s}{2.0s}[/tex]
[tex]a = 1.5 m/s^2[/tex]
Therefore, the magnitude of the acceleration of the car is 1.5 m/s².
Option D) 1.5 m/s² is the correct answer.
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How is Impulse related to Force?