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1 year ago

How many and what type of receptacles are connected to this circuit?

1 answer:

7 0

Answer:They are two types of receptacles used in this (A 16) circuit. i.) Receptacles at 120W: 5 Numbers ii.) Weatherproof receptacles at 120W: 1 Number!

Explanation:

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What is the maximal coefficient of performance of a refrigerator which cools down 10 kg of water (and then ice) to -6C. Upper he

inysia [295]

Given:

Temperature of water, $T_{1}$ = $-6^{\circ}C$ =273 +(-6) =267 K

Temperature surrounding refrigerator, $T_{2}$ = $21^{\circ}C$ =273 + 21 =294 K

Specific heat given for water, $C_{w}$ = 4.19 KJ/kg/K

Specific heat given for ice, $C_{ice}$ = 2.1 KJ/kg/K

Latent heat of fusion, $L_{fusion}$ = 335KJ/kg

Solution:

Coefficient of Performance (COP) for refrigerator is given by:

Max $COP_{refrigerator}$ = $\frac{T_{2}}{T_{2} - T_{1}}$

= $\frac{267}{294 - 267}$ = 9.89

Coefficient of Performance (COP) for heat pump is given by:

Max $COP_{heat pump}$ = $\frac{T_{1}}{T_{2} - T_{1}}$ $\frac{294}{294 - 267}$ = 10.89

6 0

3 years ago

A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp

Volgvan

Answer:

a) 22.5number

b) 22.22 m length

Explanation:

Given data:

Bridge length = 500 m

width of bridge = 12 m

Maximum temperature = 40 degree C

minimum temperature = - 35 degree C

Maximum expansion can be determined as

$\Delta L = L \alpha (T_{max} - T_{min})$

where , \alpha is expansion coefficient $= 12\times 10^{-6}$ degree C

SO, $\Delta L = 500\times 12\times 10^{-6}\times ( 40 - (-35))$

$\Delta L = 0.45 m = 450 mm$

number of minimum expansion joints is calculated as

$n = \frac{450}{20} = 22.5$

b) length of each bridge

$Length = \frac{500}{22.5} = 22.22 m$

8 0

2 years ago

What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?

Yakvenalex [24]

Answer:

$Activation\ Energy=2.5\times 10^{-19}\ J$

Explanation:

Using the expression shown below as:

$N_v=N\times e^{-\frac {Q_v}{k\times T}$

Where,

$N_v$ is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

${Q_v}$ is the activation energy

T is the temperature

Given that:

$N_v=2.3\times 10^{13}$

N = 10 moles

1 mole = $6.023\times 10^{23}$

So,

N = $10\times 6.023\times 10^{23}=6.023\times 10^{24}$

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (425 + 273.15) K = 698.15 K

T = 698.15 K

Applying the values as:

$2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$Q_v=2.5\times 10^{-19}\ J$

4 0

2 years ago

A company buys a machine for $12,000, which it agrees to pay for in five equal annual payments, beginning one year after the dat

Yuki888 [10]

Answer:

$7,778.35

Explanation:

At year 3, the final payment of the remaining balance is equal to the present worth P of the last three payments.

First, calculate the uniform payments A:

A = 12000(A/P, 4%, 5)

= 12000(0.2246) = 2695.2 (from the calculator)

Then take the last three payments as its own cash flow.

To calculate the new P:

P = 2695.2 + 2695.2(P/A, 4%, 2) = 2695.2 + 2695.2(1.886) = 7778.35

Therefore, the final payment is $7,778.35

4 0

3 years ago

Find the mechanical average of a wheel axle System of the wheel has a radius of 1.5 feet in the accident has a radius of 6 inche

kvasek [131]

Answer:

Mechanical average of a wheel = 3

Explanation:

Given:

Radius of wheel = 1.5 ft = 1.5 x 12 = 18 inches

Radius of axle = 6 inches

Find:

Mechanical average of a wheel

Computation:

Mechanical average of a wheel = Radius of wheel / Radius of axle

Mechanical average of a wheel = 18 / 6

Mechanical average of a wheel = 3

4 0

3 years ago