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WARRIOR [948]
3 years ago
6

A string of ASCII characters has been converted to hexadecimal resulting in the following message: 4A EF 62 73 73 F4 E5 76 E5 Of

the eight bits in each pair of digits, the leftmost bit is a parity bit. (a) Convert the string to the bit form (b) Decode the message (c) Which parity code was used, even or odd?
Engineering
1 answer:
Dafna11 [192]3 years ago
5 0

Answer:sorry I don’t know

Explanation:

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1.The HCF of 15 and 20?<br>2.5!=??...????​
nydimaria [60]

Answer:

the h.c.f of 15and20 is 5

7 0
3 years ago
Read 2 more answers
1)Saturated steam at 1.20bar (absolute)is condensed on the outside ofahorizontal steel pipe with an inside and outside diameter
S_A_V [24]

Answer:

hi = 7026.8  W/m^2.k

Explanation:

Given data :

pressure of saturated steam = 1.2 bar

Horizontal steel pipe : inside diameter = 0.620 , outside diameter = 0.750 inches

temperature of water at entry = 60°F

temperature of water at exit = 75°F

velocity of water = 6 ft/s

<u>Calculate the Inside convective heat transfer coefficient ( hi ) </u>

mean temperature ( Tm ) = 60 + 75 / 2 = 67.5°F ≈ 292.877 K ≈ 19.727°C

next : find the properties of water at this temperature ( 19.727°C )

thermal conductivity = 0.598  w/m.k

density = 1000 kg/m^3

specific heat ( Cp ) = 4.18 KJ/kg.k

viscosity = 0.001 pa.s

velocity of water = 6 ft/s ≈ 1.8288 m/s

∴ Re ( Reynolds number ) = 28712.16

and Prandtl number ( Pr ) = (4180 * 0.001) / 0.598  = 6.989

finally<em> to determine the inside convective heat transfer coefficient we will apply the Dittos - Bolter equation</em>

hi = 7026.8 w/m^2.k

attached below is the remaining solution

3 0
3 years ago
Assume the transistor is biased in the saturation region at VGS 4 V. (a) Calculate the ideal cutoff frequency. (b) Assume that t
insens350 [35]

Answer:

hello your question is incomplete attached below is the complete question and the detailed solution

Answer: A) 5.17 GHz

              B) 1.01 GHz

Explanation:

Assuming the transistor is biased and considering the two conditions as given in A and B attached below is a detailed solution to the given problem

4 0
3 years ago
A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one v
In-s [12.5K]

Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

6 0
3 years ago
Meeeeep
lesya [120]

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5 0
3 years ago
Read 2 more answers
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