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3 years ago

_____is audio that has been changed into patterns of numbers that can't be heard.

Engineering

2 answers:

Flura [38]3 years ago

8 0

Answer:

A. Digital audio

Explanation:

In digital audio, the sound wave of the audio signal is encoded as numerical samples in a continuous sequence.

Alla [95]3 years ago

7 0

Answer:

I think the answer is A. Digital audio. Hope it is right and it helps:)

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A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t

Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

6 0

3 years ago

B) State the essential difference between a plain carbon steel and an alloy steel

choli [55]

Answer:

Plain carbon steel has no or trace external elements while alloy steel has high amount of other elements.

Explanation:

Plain carbon steel has no or trace amount of other elements while alloy steel has high amount of other elements in their composition.

The presence of other elements in alloy steel improvise several physical properties of the steel while plain carbon steel has the basic properties.

6 0

3 years ago

A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the com

Rudiy27

Answer:

Hello your question has some missing information below are the missing information

The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump

answer : 2.49

Explanation:

For vapor-compression refrigeration cycle

P1 = P4 ; P1 = 140 kPa

P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa

From pressure table of R 134a refrigerant

h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg

h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )

= 296.8kJ/kg

h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg

also h4 = 95.47 kJ/kg

To determine the coefficient of performance

Cop = ( h1 - h4 ) / ( h2 - h1 )

∴ Cop = 2.49

3 0

3 years ago

Use the graph to determine which statement is true about the end behavior of f(x).

Airida [17]

Answer:

As the x-values go to negative infinity, the function’s values go to positive infinity.

Explanation:

if the ans choices are:

As the x-values go to negative infinity, the function’s values go to negative infinity.

As the x-values go to negative infinity, the function’s values go to positive infinity.

As the x-values go to positive infinity, the function’s values go to negative infinity.

As the x-values go to positive infinity, the function’s values go to zero.

the ans is the 2nd choice

4 0

3 years ago

Read 2 more answers

A rigid tank contains 2 kg of N2 and 4 kg of Co2 at temperature of 25 C and 1 MPa. Find the partial pressure of each gas respect

lions [1.4K]

Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.

Explanation: Dalton's Law of Partial Pressure states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:

$P_{total} = P_{1}+P_{2}+...$

The proportion of each individual gas in the total pressure is expressed in terms of mole fraction:

$X_{i}$ = moles of a gas / total number moles of gas

The rigid tank has total pressure of 1MPa.

Nitrogen gas:

molar mass = 14g/mol

mass in the tank = 2000g

number of moles in the tank: $n=\frac{2000}{14}$ = 142.85mols

Carbon Dioxide:

molar mass = 44g/mol

mass in the tank = 4000g

number of moles in the tank: $n=\frac{4000}{44}$ = 90.91mols

Total number of moles: 142.85 + 90.91 = 233.76 mols

To calculate partial pressure:

$P_{i}=P_{total}.X_{i}$

For Nitrogen gas:

$P_{N_{2}}=1.\frac{142.85}{233.76}$

$P_{N_{2}}$ = 0.6

For Carbon Dioxide:

$P_{total}=P_{N_{2}}+P_{CO_{2}}$

$P_{CO_{2}} = P_{total}-P_{N_{2}}$

$P_{CO_{2}}=1-0.6$

$P_{CO_{2}}=$ 0.4

Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.

4 0

2 years ago