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2 years ago

What type of cable communicates binary data by changing the voltage between two ranges?

1 answer:

6 0

The sending device communicates binary data across these copper wires by changing the voltage between two ranges. The system at the receiving end is able to interpret these voltage changes as binary ones and zeros which can then be translated into different forms of data.

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If the electric field just outside a thin conducting sheet is equal to 1.5 N/C, determine the surface charge density on the cond

Dennis_Churaev [7]

Answer:

The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²

Explanation:

The electric field intensity due to a thin conducting sheet is given by the following formula:

Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)

From this formula:

Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)

We have the following data:

Electric Field Intensity = 1.5 N/C

Permittivity of free space = 8.85 x 10^-12 C²/N.m²

Therefore,

Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)

<u>Surface Charge Density = 26.55 x 10^-12 C/m²</u>

Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².

7 0

2 years ago

A sample of normally consolidated clay was subjected to a CU triaxial compression test that was carried out until the specimen f

QveST [7]

Answer:

Check the explanation

Explanation:

Given

1) CU traixial compression test

2) Devatoric stress at failure = бd = 50 kN/ $m^2$

3) Confining pressure at failure = бd = 48 kN/ $m^2$

4) Pore pressure at failure = u = 18 kN/ $m^2$

5) Unconfined compression stress = q = 20 kN/ $m^2$

6) Undrained cohesion = q/2 = 20 kN/ $m^2$

To find:

1) Effective and total stress strength failure envelope

Kindly check the attached image below .

7 0

2 years ago

In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe

KIM [24]

Answer:

$\eta$ =0.60

Explanation:

Given :Take $\gamma$ =1.4 for air

$P_1=100 KPa ,T_1=300K$

$\frac{V_1}{V_2}$ =r ⇒ r=16

As we know that

$T_2=T_1(r^{\gamma-1})$

So $T_2=300\times 16^{\gamma-1}$

$T_2$ =909.42K

Now find the cut off ration $\rho$

$\rho=\frac{V_3}{V_2}$

$\frac{V_3}{V_2}=\frac{T_3}{T_2}$

$\rho=\frac{2031}{909.42}$

$\rho=2.23$

So efficiency of diesel engine

$\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}$

Now by putting the all values

$\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}$

So $\eta$ =0.60

So the efficiency of diesel engine=0.60

7 0

2 years ago

A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process t

iren2701 [21]

Work done = -19.7 KJ

Heat transferred = 17.4 KJ

Explanation:

Given-

Temperature, T = 27°C

Volume, V = 0.2 m³

Pressure, $P_{1}$ = 1 bar

$v_{2}$ = 4 bar

pV¹°¹ = constant

From superheated propane table, at $P_{1}$ = 1 bar and $T_{1}$ = 27⁰C

$v_{1}$ = 0.557 m³/kg

$v_{2}$ = 473.73 KJ/kg

(a) Work = ?

We know,

V1¹°¹ = p2V2¹°¹

$V2 = (\frac{P1}{P2})^\frac{1}{1.1} * V1 \\\\V2 = \frac{1}{4}^\frac{1}{1.1} * 0.557 \\\\V2 = 0.158 m^3/kg$

At = 4 bar and v = 0.158 m³/kg

u2 = 548.45K J/kg

To find work done in the process:

$W = \frac{P2V2 - P1V1}{1-n} \\\\W = \frac{m(P2V2 - P1V1)}{1-n} \\\\W = \frac{v}{u} * \frac{P2V2 - P1V1}{1-n}\\ \\W = \frac{0.2}{0.5571} * \frac{4 X 0.158 - 1 X 0.577}{1-1.1} X 10^5 \frac{Pa}{Bar} \frac{1KJ}{10^3Nm} \\ \\W = -19.75KJ$

(b) Heat transfer = ?

$Q = m(u2 - u1) + W\\\\Q = \frac{0.2}{0.5571} * (548.45 - 473.73) + (-19.7)\\\\Q = 17.4KJ$

8 0

2 years ago

A bankrupt chemical firm has been taken over by new management. On the property they found a 20,000-m3 brine pond containing 25,

Liula [17]

Answer:

Flow-rate = 0.0025 m^3/s

Explanation:

We need to assume that the flow-rate of pure water entering the pond is the same as the flow-rate of brine leaving the pond, in other words, the volume of liquid in the pond stays constant at 20,000 m^3. Using the previous assumption we can calculate the flow rate entering or leaving the tank (they are the same) building a separable differential equation dQ/dt, where Q is the milligrams (mg) of salt in a given time t, to find a solution to our problem we build a differential equation as follow:

dQ/dt = -(Q/20,000)*r where r is the flow rate in m^3/s

what we pose with this equation is that the variable rate at which the salt leaves the pond (salt leaving over time) is equal to the concentration (amount of salt per unit of volume of liquid at a given time) times the constant rate at which the liquid leaves the tank, the minus sign in the equation is because this is the rate at which salt leaves the pond.

Rearranging the equation we get dQ/Q = -(r/20000) dt then integrating in both sides ∫dQ/Q = -∫(r/20000) dt and solving ln(Q) = -(r/20000)*t + C where C is a constant (initial value) result of solving the integrals. Please note that the integral of dQ/Q is ln(Q) and r/20000 is a constant, therefore, the integral of dt is t.

To find the initial value (C) we evaluate the integrated equation for t = 0, therefore, ln(Q) = C, because at time zero we have a concentration of 25000 mg/L = 250000000 mg/m^3 and Q is equal to the concentration of salt (mg/m^3) by the amount of liquid (always 20000 m^3) -> Q = 250000000 mg/m^3 * 20000 m^3 = 5*10^11 mg -> C = ln(5*10^11) = 26.9378. Now the equation is ln(Q) = -(r/20000)*t + 26.9378, the only thing missing is to find the constant flow rate (r) required to reduce the salt concentration in the pond to 500 mg/L = 500000 mg/m^3 within one year (equivalent to 31536000 seconds), to do so we need to find the Q we want in one year, that is Q = 500000 mg/m^3 * 20000 m^3 = 1*10^10 mg, therefore, ln(1*10^10) = -(r/20000)*31536000 + 26.9378 solving for r -> r = 0.002481 m^3/s that is approximately 0.0025 m^3/s.

Note:

ln() refer to natural logarithm
The amount of liquid in the tank never changes because the flow-rate-in is the same as the flow-rate-out
When solving the differential equation we calculated the flow-rate-out and we were asked for the flow-rate-in but because they are the same we could solve the problem
During the solving process, we always converted units to m^3 and seconds because we were asked to give the answer in m^3/seg

7 0

2 years ago