1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
love history [14]
3 years ago
7

A homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time. Determine the average cost per k

Wh for the month using the following residential rate schedule: Base monthly charge of $10.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 300 kWh per month at 6 cents/kWh.
Engineering
1 answer:
Llana [10]3 years ago
7 0

Answer:

16.2 cents

Explanation:

Given that a homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time.

Where Base monthly charge of $10.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 300 kWh per month at 6 cents/kWh.

For the first 100 kWh:

16 cent × 100 = 1600 cents = 16 dollars

Since 1 dollar = 100 cents

For the remaining energy:

260 - 100 = 160 kwh

10 cents × 160 = 1600 cents = 16 dollars

The total cost = 10 + 16 + 16 = 42 dollars

Note that the base monthly of 10 dollars is added.

The cost of 260 kWh of energy consumption in July is 42 dollars

To determine the average cost per kWh for the month of July, divide the total cost by the total energy consumed.

That is, 42 / 260 = 0.1615 dollars

Convert it to cents by multiplying the result by 100.

0.1615 × 100 = 16.15 cents

Approximately 16.2 cents

You might be interested in
Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read
damaskus [11]

Answer:

P_2-P_1=27209h

Explanation:

For pressure gage we can determine this by saying:

The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

P_1+p_1gh_1=p_2_gh_2+P_2

We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h

P_1-P_2=p_1gh_1-p_2gh_2=(55)\cdot{32.2}h-845\cdot{32.2}h

P_2-P_1=27209h

3 0
3 years ago
A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an
Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = 1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}

Cyclone is to 80 % efficient

So particulate remaining = 0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0
3 years ago
B)
Triss [41]

Answer:

2.5 is the required details

8 0
3 years ago
Race car is accelerating and has a velocity of 10 m/s @ t=0. It completes a lap on a circular track of 400 m in 14 seconds. Calc
wariber [46]

Answer:

component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²

magnitude of acceleration is  22.98 m/s²

Explanation:

given data

velocity = 10 m/s

initial time to = 0

distance s = 400 m

time t = 14 s

to find out

components and magnitude of acceleration after the car has travelled 200 m

solution

first we find the radius of circular track that is

we know  distance S = 2πR

400 = 2πR

R = 63.66 m

and tangential acceleration is

S = ut + 0.5 ×at²

here u is initial speed and t is time and S is distance

400 = 10 × 14  + 0.5 ×a (14)²

a = 3.37 m/s²

and here tangential acceleration is constant

so  velocity at distance 200 m

v² - u² = 2 a S

v² = 10² + 2 ( 3.37) 200

v = 38.05 m/s

so radial acceleration at distance 200 m

ar = \frac{v^2}{R}

ar = \frac{38.05^2}{63.66}

ar = 22.74 m/s²

so magnitude of total acceleration is

A = \sqrt{a^2 + ar^2}

A = \sqrt{3.37^2 + 22.74^2}

A = 22.98 m/s²

so magnitude of acceleration is  22.98 m/s²

8 0
3 years ago
The voltage and current at the terminals of the circuit element in Fig. 1.5 are zero fort &lt; 0. Fort 2 0 they areV =75 ~75e-10
masya89 [10]

Answer:

maximum value of the power delivered to the circuit =3.75W

energy delivered to the element = 3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750

Explanation:

V =75 - 75e-1000t V

l = 50e -IOOOt mA

power = IV = 50 * 10^-3 e -IOOOt * (75 - 75e-1000t)

=50 * 10^-3 e -IOOOt *75 (1 - e-1000t)

=

maximum value of the power delivered to the circuit =3.75W

the total energy delivered to the element = \int\limits^t_0  {3.75(e^{ -IOOOt} - e ^{-2OOOt} )} , dx \\\\

3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750

5 0
3 years ago
Other questions:
  • 4. Three routes connect an origin and a destination with performance functions tl = 8 + 0.5x1, t2 = 1 + 2x2, and t3 = 3 + 0.75x3
    9·1 answer
  • A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar abs
    8·1 answer
  • How far do you jog each morning? You prefer to jog in different locations each day and do not have a pedometer to measure your d
    14·1 answer
  • What are the BENEFITS and RISKS of using automobiles?
    7·1 answer
  • A structural component in the shape of a flat plate 29.6 mm thick is to be fabricated from a metal alloy for which the yield str
    11·1 answer
  • A long rod of 60-mm diameter and thermophysical properties rho=8000 kg/m^3, c=500J/kgK, and k=50 W/mK is initally at a uniform t
    8·1 answer
  • A landowner and a contractor entered into a written contract under which the contractor agreed to build a building and pave an a
    14·1 answer
  • Please help on two I will give brainiest​
    13·2 answers
  • One reason the shuttle turns on its back after liftoff is to give the pilot a view of the horizon. Why might this be useful?
    6·2 answers
  • The project's criteria.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!