In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of line has resistance R. For a power line that supplies power to 10,000 households, we can conclude that:________

Answer:

b. What is the operator's efficiency for the day?

                                                      AND

e. What would be the proper piece rate (rate expressed in money) for this job, assuming that the above time standard is correct?

Explanation:

Answer:

Creating an app is both an expression of our self and a reflection of what we see is missing in the world. We find ourselves digging deep into who we are, what we would enjoy working on, and what needs still need to be fulfilled. Generating an app idea for the first time can be extremely daunting. Especially with an endless amount of possibilities such as building a church app.

The uncertainty has always spawned a certain fear inside creators. The fear of creating something no one will enjoy. Spending hundreds of dollars and hours building something which might not bring back any real tangible results. The fear of losing our investment to a poor concept is daunting but not random. But simple app ideas are actually pretty easy to come by.

Great app idea generation is not a gift given to a selected few, instead, it is a process by which any of us are able to carefully explore step by step methods to find our own solution to any problem. Whether you are a seasoned creator or a novice, we have provided a few recommendations to challenge and aid you as you create your next masterpiece.

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Answer:

B) gate-source junction is reverse-biased

Explanation:

FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow"   in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".

In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".

Answer:

[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy at constant pressure:

[tex]ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}[/tex]

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

[tex]\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,[/tex]

We obtain the expression to compute the specific entropy change:

[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]

Best regards.

Answer:

Explanation:

Temperature of atmospheric air To = 25°C = 298 K

Free  stream velocity of air Vo = 25 m/s

Length and width of plate = 1m

Temperature of plate Tp = 125°C = 398 K

We know for air, Prandtl number Pr = 1

And for air, thermal conductivity K = 24.1×10?³ W/mK

Here, charectorestic dimension D = 1m

Given value of Reynolds number Re = 105

For laminar boundary layer flow over flat plate

= 3.402

Therefore, hx = 0.08199 W/m²K

So, heat transfer rate q = hx×A×(Tp – To)

                                          = 0.08199×1×(398 – 298)

Answer:

a) Mt = 0.0023229

b) = U1 = 214.07

c) = V₁  = 0.861 m³/kg

d) = Vr1 = 621.2

Explanation:

Given that

R = 0.287 KJ/kg.K, T1 = 300 K , P1 = 100 kPa , V1 = 500 cm³, r = 10 , Q = 60 kW , Speed N = 5600 RPM, Number of cylinders K = 4

specific heat at constant volume Cv = 0.7174 kJ/kg.K

Specific heat at constant pressure is 1.0045 Kj/kg.K

a)  To determine the total mass (kg) of air in the engine.

we say

P1V1 = mRT1

we the figures substitute

(100 x 10³) ( 500 x  10⁻⁶) = m ( 0.287 x  10³) ( 300 )

50 = m x 86100

m = 0.00005 / 86100 = 0.0005807 ( mass of one cylinder)

Total mass of 4 cylinder

Mt = m x k

Mt = 0.0005807 x 4

Mt = 0.0023229

b) To determine the specific internal energy (kJ/kg) at state 1

i.e at T1 = 300

we obtain the value of specific internal energy U1 at 300 K ( state 1) from the table ideal gas properties of air.

U1 = 214.07

c) To determine the specific volume (m³/kg) at state 1.

we say

V₁ = V1/m

V₁ = (500 x  10⁻⁶) / 0.0005807

V₁  = 0.861 m³/kg

d) To determine the relative specific volume at state 1.

To obtain the value of relative specific volume at 300 K ( i.e state 1) from the table ideal gas properties of air.

At T1 = 300 k

Vr1 = 621.2

Answer:

Move the buret clamp to a ring stand with a larger base.

Explanation:

A right stand is used for titration experiments in the laboratory. It holds the burette firmly during experiments so that accurate readings can be taken.

The right stand is made up of support base, vertical stainless steel, clamp with adjustable screw that holds on to the vertical rod.

The clamp is used to hold the burette in place.

If after clamping a buret to a ring stand, you notice that the set-up is tippy and unstable, the best action will be to move the buret clamp to a ring stand with a larger base.

The larger base provides a better center of gravity and stabilises the setup

Answer:

New electric field = 18 N/C

Explanation:

Given:

Length (E1) = 2 cm

New length (E2) = 4 cm

Electric field =  36 N/C

Find:

New electric field

Computation:

New electric field = 36 [2 / 4]

New electric field = 36 [1/2]

New electric field = 18 N/C

Answer:

A) EXIT TEMPERATURE = 14⁰C

b) rate of heat transfer of air = - 13475.78 = - 13.5 kw

Explanation:

Given data :

diameter of duct = 20-cm = 0.2 m

length of duct = 12-m

temperature of air at inlet= 50⁰c

pressure = 1 atm

mean velocity = 7 m/s

average heat transfer coefficient = 85 w/m^2⁰c

water temperature = 5⁰c

surface temperature ( Ts) = 5⁰c

properties of air at 50⁰c and at 1 atm

= 1.092 kg/m^3

Cp = 1007 j/kg⁰c

k = 0.02735 W/m⁰c

Pr = 0.7228

v  = 1.798 * 10^-5 m^2/s

determine the exit temperature of air and the rate of heat transfer

attached below is the detailed solution

Calculate the mass flow rate

= p*Ac*Vmean

= 1.092 * 0.0314 *  7 = 0.24 kg/s