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OLga [1]
2 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]2 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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3 years ago
A student is running at her top speed of 5.4 m/s to catch a bus, which is stopped at the bus stop. When the student is still a d
olya-2409 [2.1K]

Answer:

a) t=8.19s; x=44.2m

b) v=1.401 m/s

c) see attachment

d) The second solution is a later time at which the bus catches the student. v=9.40 m/s

e) No, she won't.

f) v=3.63m/s;t=21.2; x=77m

Explanation:

a) The motion of both the bus and the student can be explained by the equation x= v_{0} +\frac{1}{2}at^{2}. Since the student is not accelerating, but rather maintaining a constant speed; the particular equation that describes the motion of the student is: x_{student} = 5.4 \frac{m}{s} * t. Meanwhile, since the bus starts its motion at an initial velocity of zero, the equation that describes its motion is: x_{bus} =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2}. The motion of the student relative to that of the bus can be described by the equation: x_{s} =x_{bus} +38.5m. By replacing terms in the last equation we end up with the following quadratic equation: (0.0855 \frac{m}{s^{2} } * t^{2} )-(5.4\frac{m}{s} *t)+38.5m =0. Solving the quadratic equation will yield two solutions; t1=8.19s and t2=55.0s. By plugging in t1 onto the equation that describes the motion of the student we will find the distance runned by her, x=44.2m.

b) The velocity of the bus can be modeled by the equation v^{2} = v_{0} ^{2} +2ax. Since the initial velocity of the bus is zero, the first term of the equation cancels. Next, we solve for v and plug in the acceleration of 0.171 m/s2 and the distance 5.74m (traveled by the bus, note: this is equal to the 44.2m travelled by the student minus the 38.5m that separated the student from the bus at the beginning of the problem).

c) The equations that make up the x-t graph are: x = 5.4 \frac{m}{s} * t and  x =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2} + 38.5; as described in part a.

d) The first solution states that the student would have to run 44.2 m in 8.19 s in order to catch the bus. But, at that point, the student has a greater speed than that of the bus. So if both were to keep he same specified motion, the student would run past the bus until it reaches a velocity greater than that of the student. At which point, the bus will start to narrow the distance with the student until it finally catches up with the student 55 seconds after both started their respective motion.

e) No, because the quadratic equation (0.0855 \frac{m}{s^{2} } * t^{2} )-(3.5\frac{m}{s} *t)+38.5m =0 has no solution. This means that the two curves that describe the distance vs time graph for both student and bus do not intersect.

f) This answer is reached by finding b of the quadratic equation. The minimum that b can be in order to find a real answer to the quadratic equation is found by solving b^{2} -4ac=0. If we take a = 0.0855 and c = 38.5, then we find that the minimum speed that the student has to run at is 3.63 m/s. If we then solve the quadratic equation (0.0855 \frac{m}{s^{2} } * t^{2} )-(3.63\frac{m}{s} *t)+38.5m =0,we will find that the time the student will run is 21.2 seconds. By pluging in that time in the equation that describes her motion: x_{student} = 3.63 \frac{m}{s} * t we find that she has to run 77 meters in order to catch the bus.

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