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Ahat [919]
2 years ago
8

How does error detection take place

Engineering
1 answer:
sattari [20]2 years ago
6 0

Answer: To detect and correct errors, additional bits are added to the data bits at the time of transmission. The additional bits are called parity bits. They allow detection or correction of errors. The data bits along with the parity bits form a code word.

Explanation:

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All circuits need three basic parts: an energy source, wires, and the object that is going to change the electrical energy into
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load every electric circuit,regardless of where it is or how large or small, has four basic parts: an energy source (ac or dc),a conductor (wire), an electrical load (device), and at least one controller(switch)
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What are the dimensions of the base of the pyramid?
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b

Explanation:

bcus it is

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with a digital system, if you have measured incorrectly and use too low of a kvp for adequate penetration, what do you need to d
Lubov Fominskaja [6]

The x-ray beam's penetrating power is regulated by kVp (beam quality). Every time an exposure is conducted, the x-rays need to be powerful (enough) to sufficiently penetrate through the target area.

<h3>How does kVp impact the exposure to digital receptors?</h3>

The radiation's penetration power and exposure to the image receptor both increase as the kVp value is raised.

<h3>Exposure to the image receptor is enhanced with an increase in kVp, right?</h3>

Due to an increase in photon quantity and penetrability, exposure at the image receptor rises by a factor of five of the change in kVp, doubling the intensity at the detector with a 15% change in kVp.

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5 0
1 year ago
Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Stea
Nataly [62]

Answer:

1) The exergy of destruction is approximately 456.93 kW

2) The reversible power output is approximately 5456.93 kW

Explanation:

1) The given parameters are;

P₁ = 8 MPa

T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

h₁ = 3399 kJ/kg

P₂ = 2 MPa

T₂ = 350°C

From which we have;

s₂ = 6.958 kJ/(kg·K)

h₂ = 3138 kJ/kg

P₃ = 2 MPa

T₃ = 500°C

From which we have;

s₃ = 7.434 kJ/(kg·K)

h₃ = 3468 kJ/kg

P₄ = 30 KPa

T₄ = 69.09 C (saturation temperature)

From which we have;

h₄ = h_{f4} + x₄×h_{fg} = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg

s₄ =  s_{f4} + x₄×s_{fg} = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, \dot X_{dest}, is given as follows;

\dot X_{dest} = T₀ × \dot S_{gen} = T₀ × \dot m × (s₄ + s₂ - s₁ - s₃)

\dot X_{dest} = T₀ × \dot W×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ \dot X_{dest} = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138  - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, \dot W_{rev} = \dot W_{} + \dot X_{dest} ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

6 0
3 years ago
Technician A says that mechanical shifting controls can wear out over time. Technician B says that vacuum control rubber diaphra
diamong [38]

Based on the information, both technician A and technician B are correct.

<h3>How to depict the information?</h3>

From the information given, Technician A says that mechanical shifting controls can wear out over time.

Technician B says that vacuum control rubber diaphragms can deteriorate over time.

In this case, both technicians are correct as the information depicted is true.

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8 0
2 years ago
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