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OLga [1]
3 years ago
12

Light emitting diode (LEDs) light bulbs have become required in recent years, but do they make financial sense? Suppose a typica

l 60-watt incandescent light bulb costs $.39 and lasts for 1,000 hours. A 15-watt LED, which provides the same light, costs $3.10 and lasts for 12,000 hours. A kilowatt hour of electricity costs $.115. A kilowatt-hour is 1,000 watts for 1 hour. However, electricity costs actually vary quite a bit depending on location and user type. An industrial user in West Virginia might pay $.04 per kilowatt-hour whereas a residential user in Hawaii might pay $.25. You require a return of 11 percent and use a light fixture 500 hours per year. What is the break-even cost per kilowatt-hour?
Physics
1 answer:
kogti [31]3 years ago
3 0

Answer:

The break even cost is $0.0063825

Explanation:

Break-even cost is the amount of money, or change in value, which equates to the amount at which an asset must be sold to equal the cost of acquiring it. For easier understanding it can be thought the amount of money for which a product or service must be sold to cover the costs of manufacturing or providing it.

Wattage = W

Cost per kilo watt hour = C

Number of hours per year = H

Price per bulb/CFL = P

Discount rate = 11%

Life of bulb = 2 years

Price of bulb = $0.39

Wattage consumption of bulb per hours = 60

Life of CFL = 24 years

Price of CFL = $3.10

Wattage consumption of CFL per hour = 15

Calculate the Equated Annual Cost (EAC) of bulb

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2years)}/ (PVIFA 11%, 2years)

PVIFA 11%, 2years = Annuity PV Factor = [1 – {(1 + r)^(-n)}]/r, where r is the rate per period and n is the number per periods

PVIFA 11%, 2 years = [1 – {(1 + 0.11)^(-2)}]/0.11 = 1.712523 (for 2 years)

PVIFA 11%, 24 years = [1 – {(1 + 0.11)^(-24)}]/0.11 = 8.348136 (for 2 years)

<u>Calculate the EAC of bulb</u>

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2 years)}/ (PVIFA 11%, 2 years)

EAC = {- 0.39 - (60/1000 x H x C) x (1.712523)}/ (1.712523)

EAC = {-0.39 – (51.37570 x C)}/ 1.712523, <em>consider this equation 1</em>

<u>Calculate the EAC of CFL</u>

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 24 years)}/ (PVIFA 11%, 24 years)

EAC = {- 3.10 - (15/1000 x 500 x C) x (8.348136)}/ (8.348136)

EAC = {-3.10 – (62.61102 x C)}/8.348137, <em>consider this equation 2</em>

<u>Equate 1 and 2 to find the amount of C</u>

{-0.39 – (51.37570 x C)}/ 1.712523 = {-3.10 – (62.61102 x C)}/8.348137

{-0.39 – (51.37570 x C) x 8.348137} = {-3.10 – (62.61102 x C) x 1.712523}

C = $0.0063825

Thus, the break- even cost per kilo – watt is $0.0063825

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1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

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Substituting t = 0.200 s, we find

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The tangential acceleration of a point rotating at a distance r from the centre of the circle is

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First of all, we need to convert the angular acceleration into rad/s^2:

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A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp
amm1812

Answer:

the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

Explanation:

Given that,

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The diffrential equation of motion with damping is given by

m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}

substitute the value of m =1kg, k = 21N/M, and \beta = 10

1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}

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suppose the equation of the form x =e^m^t,

and the auxilliary equation is given by

m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3

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since time is 0 then mass is one meter below

so x(0) = 1

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x'(0) = 0

substitute the initial condition

C_1 +C_2 = 1

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C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}

substitute for C₁ and C₂ in general solution

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