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laila [671]
3 years ago
6

You want to capture and digitize an analog signal that has a frequency bandwidth of 5500 Hz. You select a sampling rate of 5500

samples per second. What happens to your digitized signal? a. You should be fine since your sampling rate is the same as the analog signal’s bandwidth b. You cannot sample an analog signal since it is infinite and continuous c. Your digital signal will experience aliasing because you did not follow Nyquist sampling rate requirements d. None of the above
Physics
1 answer:
VladimirAG [237]3 years ago
6 0

Answer:

c.

Explanation:

When you take samples of an analog signal, you must take samples fast enough so then you can recover the original signal, just passing the digitized signal through a low-pass filter.

The Nyquist criteria states that in order to be able to recover the original signal completely, you must take samples at a rate greater than 2 times the highest frequency component of the signal.

In our case, we should sample the analog signal at a rate > 2*5500 Hz = 11000 Hz.

If we sampled the 5500 Hz signal (assuming be a sinusoid just for simplicity) two times each cycle, we could be so unfortunate that the samples fall exactly when the signal crosses by zero, so the digitized signal would be only a train of zeros.

So, if we sample the signal only one time in each cycle, clearly we will not  be able  to  recover the signal, and the digital signal will experience aliasing, due we are not following Nyquist sample rate  requirements, as stated  in the  option c.

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An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dro
kupik [55]

Answer:

a) Q_{in} = 13.742\,kW, b) \Delta S = 370.15\,\frac{kJ}{K}

Explanation:

a) The heat transfered to the egg is computed by the First Law of Thermodynamics:

Q_{in} +U_{sys,1} - U_{sys,2} = 0

Q_{in} = U_{sys,2} - U_{sys,1}

Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})

Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)

Q_{in} = 13.742\,kW

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

\Delta S = \frac{Q_{in}}{T_{in}}

\Delta S = \frac{13.742\,kJ}{370.15\,K}

\Delta S = 370.15\,\frac{kJ}{K}

3 0
3 years ago
A 19.12 g mixture of Ca(NO3)2 and KCl is dissolved in 149 g of water. The freezing point of the solution was measured as −5.77 ∘
hichkok12 [17]

Answer:

The mass percentage of calcium nitrate is 31.23%.

Explanation:

Let the the mass of calcium nitrate be x and mass of potassium chloride be y.

Total mass of  mixture = 19.12 g

x + y = 19.12 g..(1)

Mass of solvent = 149 g = 0.149 kg

Freezing point of the solution,T_f = -5.77 °C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:

i = 3

i' = 2

Freezing point of water = T = 0°C

\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC

\Delta T_f=i\times K_f\times m

Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}

5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})

On solving we get:

\frac{3x}{164 g/mol}+\frac{2x}{74.5 g/mol}=0.4622 mol....(2)

Solving equation (1)(2) for x and y:

x =5.973 g

y = 13.147 g

Mass percent of Ca(NO_3)_2 in the mixture:

\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%

The mass percentage of calcium nitrate is 31.23%.

5 0
3 years ago
The specific heat capacity of ethanol is 2440J/kg °C. How many joules of energy will be required to heat 150g ethanol to 35°C if
m_a_m_a [10]

Answer:

Required energy = 4758 J

Explanation:

Specific heat capacity of a material is the amount of energy required to raise the temperature of one kilogram (kg) of that material through one degree Celsius (°C).    

Given data :

Specific heat capacity = c = 2440 J/kg.°C

Mass = m = 150 g = 0.15 kg

Initial temperature = 22°C

Final temperature = 35°C

Change in Temperature = ΔT = 13°C

Energy = E = ?

Using the following formula and substituting the values, we get:

E = m × c × ΔT

E = 0.15 × 2440 × 13

E = 4758 J

8 0
3 years ago
A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l
baherus [9]

Answer:

(a) The density of the object is 316/343 × the density of the oil

(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume

Explanation:

(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

V = \dfrac{\pi r^{2} h}{3}

The volume of the object is therefore;

\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3

Where 6 cm is above the oil level we have;

\dfrac{\pi \times \left (6 \times \dfrac{4}{14}   \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3 above the oil level

Therefore, volume of the oil displaced = 68\tfrac{116}{147}\pi cm³ = 216.11 cm³

The density of the object is thus;

\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times  Density \ of \ the \ oil = \dfrac{316}{343}  \right ) \times  Density \ of \ the \ oil

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = 2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3

The fraction of the volume displaced, x, after immersing the object is given as follows;

x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

8 0
3 years ago
A bag is dropped from a hovering helicopter. When the bag has fallen 2 seconds,
denis-greek [22]
A) Vt = Vo - gt
= 0 - 9.8 . 2
= - 19.6 m/sec (acting downward)
b) ∆y = Vt^2 - Vo^2 / 2g
= (-19.6)^2 - 0 / 19.6 = 19.6 meters
5 0
2 years ago
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