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3 years ago

15

In a team, a person’s efforts are less identifiable than when that person works independently. Because the person’s efforts are

less identifiable, in extreme circumstances this can lead to __________.

1 answer:

Eduardwww [97]3 years ago

3 0

The answer is deindividuation - a psychological state in which a person does not feel individual responsibility.

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Hwan also wants to determine the depreciation balance for the first year and the last year of the useful life of the product har

Elden [556K]

In Menlo Park, California, Hwan Rhee is debating whether to launch ChargeAll, a software firm that will create full-room wireless chargers for any kind of mobile electronic device.

In Menlo Park, California, Hwan Rhee is debating whether to launch ChargeAll, a software firm that will create full-room wireless chargers for any kind of mobile electronic device. In order to examine the financial information for a startup loan that would pay for the components and product manufacture, Hwan is using an Excel worksheet. He requests your assistance in fixing mistakes and performing financial computations in the worksheet.

Visit the worksheet for loan analysis. Hwan requests that you fix the mistakes in the worksheet before he can compute the principal and interest payments on the loan.

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4 0

1 year ago

A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid.

Trava [24]

Answer:

a. Heat removal rate will increase

b. Heat removal rate will decrease

Explanation:

Given that

One end of rod is connected to the furnace and rod is long.So this rod can be treated as infinite long fin.

We know that heat transfer in fin given as follows

$Q_{fin}=\sqrt{hPKA}\ \Delta T$

We know that area

$A=\dfrac{\pi}{4}d^2$

Now when diameter will triples then :

$A_f=\dfrac{\pi}{4}{\left (3d \right )}^2$

$A_f=9A$

$Q'_{fin}=\sqrt{9hPKA}\ \Delta T$

$Q'_{fin}=3\sqrt{hPKA}\ \Delta T$

$Q'_{fin}=3Q$

So the new heat transfer will increase by 3 times.

Now when copper rod will replace by aluminium rod :

As we know that thermal conductivity(K) of Aluminium is low as compare to Copper .It means that heat transfer will decreases.

3 0

3 years ago

The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevat

ankoles [38]

Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

$\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}$

Properties: The density of water $\delta = 1000 kg/m^3$

Conversions:

$165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\$

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

$e_{mech_{in}} - e_{mech_{out}} = gh - 0$

Then;

$gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}$

gh = 0.491 kJ/kg

$\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg$

= 1559 kW

Therefore; the overall efficiency is:

$\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}$

$= \dfrac{1166 \ kW}{1559 \ kW}$

= 0.75

= 75%

b) mechanical efficiency of the turbine:

$\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}$

thus;

$\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%$

6 0

3 years ago

You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are current

postnew [5]

Answer:

$D=0.016m$

Explanation:

From the question we are told that:

Discharge Rate $F_r=0.5kgls$

Pressure $P=15Kpa$

Temperature $T=25=>298K$

Ambient pressure is 1 atm.

Generally the equation for Density is mathematically given by

$\rho=\frac{PM}{RT}$

$\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}$

$\rho=16.958kg/m^2$

Generally the equation for Flow rate is mathematically given by

$F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}$

Where

$Q=Heat coefficient\ ratio\ of\ Nitrogen$

$Q=1.4$

$\mu= Discharge\ coefficient$

$\mu=0.68$

Therefore

$0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}$

$A=2.129*10^{-4}$

Where

$A=\frac{\pi}{4}D^2$

$\frac{\pi}{4}D^2=2.129*10^{-4}$

$D=0.016m$

8 0

3 years ago

A wastewater treatment plant discharges 2.0 m^3/s of effluent having an ultimate BOD of 40.0 mg/L into a stream flowing at 15.0

dexar [7]

Answer:

What grade is this is for??

4 0

3 years ago