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pickupchik [31]
2 years ago
12

A digital-to-analog converter uses a reference voltage of 120 V dc and has eight binary digit precision. In one of the sampling

instants, the data contained in the binary register = 01010101. If a zero order hold is used to generate the output signal, determine the voltage level of the signal.
Engineering
2 answers:
Pavel [41]2 years ago
8 0

Answer: Vo= 39.84 volts

Explanation:

Let Vo to be the voltage level of the signal,

Therefore,

Vo = 120{0.5(0) + 0.25(1) + 0.125(0) + 0.0625 (1) + 0.03125(0)

+ 0.015625(1) + 0.007812(0) +0.003906(1)}

Vo= 39.84 volts

Digiron [165]2 years ago
3 0

Given Information:  

Reference voltage = Vr = 120 Volts

Number of bits = 8

Required Information:  

Analog voltage at 01010101 = V = ?

Answer:

V = 39.84372 Volts

Explanation:

Binary Register: 0 1 0 1 0 1 0 1

V = Vr*[ 2⁻¹(0) + 2⁻²(1) + 2⁻³(0) + 2⁻⁴(1) + 2⁻⁵(0) + 2⁻⁶(1) + 2⁻⁷(0) + 2⁻⁸(1)]

V = 120*[ 0.5(0) + 0.25(1) + 0.125(0) + 0.0625(1) + 0.03125(0) + 0.015625(1) + 0.007812(0) + 0.003906(1)]

V = 120*[ 0.25 + 0.0625 + 0.015625 + 0.003906]

V = 120*[ 0.332031]

V = 39.84372 Volts

Therefore, the analog voltage corresponding to binary register 01010101 is 39.84372 Volts

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k = 1.91 × 10^-5 N m rad^-1

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2 years ago
Two kilograms of oxygen fills the cylinder of a piston-cylinder assembly. The initial volume and pressure are 2 m3 and 1 bar, re
valentinak56 [21]

Answer: Heat transfer (Q) is 521 kJ.

Explanation: In the piston-cilinder assembly, we can suppose that the oxygen act as an ideal gas, so, it can be used the General Gas Equation:

PV=\frac{m}{M}RT, where:

P is pressure;

V is volume;

m is mass;

M is molar mass;

R is a constant: R = 8.314.10^{-5} m³bar.K⁻¹.mol⁻¹;

T is temperature;

Using this equation, find the intial temperature:

PV = \frac{m}{M}RT

1.2 = \frac{2}{16}.8.314.10^{-5}.T

T = 1.924.10^{5} K

To determine the final temperature, use Combined Gas Law:

\frac{P_{i} . V_{i} }{T_{i} } = \frac{P.V }{T}, in which, the left side of the equality is related to the initial values and the right side, to the final values.

As pressure is constant:

\frac{V_{i} }{T_{i} } =\frac{V}{T}

T = \frac{V.T_{i} }{V_{i} }

T = \frac{4.1.924.10^{-5} }{2}

T = 3.85.10^{5} K

With the temperatures, calculate the heat transfer of the process:

Q = m.k.ΔT, where:

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Q = m.k.ΔT

Q = 2.1.35.(3.85 - 1.92).10^{5}

Q = 521 kJ

The heat transfer in the process is 521 kJ.

6 0
2 years ago
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Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the o
tresset_1 [31]

Answer:

Vx = 6.242 x 10raised to power 15

Vy = -6.242 x 10raised to power 15

Explanation:

from E = IVt

but V = IR from ohm's law and Q = It from faraday's first law

I = Q/t

E = Q/t x V x t = QV

hence, E =QV

V = E/Q

3 0
2 years ago
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A plate of an alloy steel has a plane-strain fracture toughness of 50 MPa√m. If it is known that the largest surface crack is 0.
Ivahew [28]

Answer:

option B is correct. Fracture will definitely not occur

Explanation:

The formula for fracture toughness is given by;

K_ic = σY√πa

Where,

σ is the applied stress

Y is the dimensionless parameter

a is the crack length.

Let's make σ the subject

So,

σ = [K_ic/Y√πa]

Plugging in the relevant values;

σ = [50/(1.1√π*(0.5 x 10^(-3))]

σ = 1147 MPa

Thus, the material can withstand a stress of 1147 MPa

So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.

8 0
2 years ago
The extruder head in a fused- deposition modeling setup has a diameter of 1.25 mm (0.05 in) and produces layers that are 0.25mm
Angelina_Jolie [31]

Answer:

The time taken will be "1 hour 51 min". The further explanation is given below.

Explanation:

The given values are:

Number of required layers:

= \frac{38}{0.25}

= 152 \ layers

Diameter (d):

= 1.25 mm

Velocity (v):

= 40 mm/s

Now,

The area of one layer will be:

= 38\times 38 \ mm^2

= 1444 \ mm^2

The area covered every \second will be:

= d\times v

= 1.25\times 40

= 50 \ mm^2

The time required to deposit one layer will be:

= \frac{1444}{50}

= 28.88 \ sec

The time required for one layer will be:

= 15 \ sec

∴ Total times required for one layer will be:

= 15+28.88

= 43.88 \ sec

So,

Number of layers = 152

Therefore,

Total time will be:

= 152\times 43.88

= 6669.76 \ sec

= 1 \ hour \ 51 \ min

6 0
2 years ago
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