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kumpel [21]
3 years ago
15

LED light bulbs are more energy efficient, last longer, cost more, and have a smaller environmental impact from manufacture to r

ecycling or disposal. The question is which costs less: incandescent, compact fluorescent (CFL), or LED lights?
(a) Ignore the time value of money. What is the average cost per hour for 25,000 hours of light-mg in a community with exist0.15/kWh electricity: for each choice?
(b) If the light is on 5%, 30%, or 60% of the time, how long until does it take for the 25,000 hours?
(c) Which type of light should a homeowner buy?
Physics
1 answer:
Nadusha1986 [10]3 years ago
5 0

Answer:

a) Cost_incandescent  = $ 375 , Cost _CFL = $ 75 , Cost_LED = $ 45

b)  total = 20833 days ,  total_day = 3472 days ,  Total_days = 1736 d

c) LED

Explanation:

This problem is interesting to solve it, we must have a very important factor, such as the equivalence of the light emitted by the three systems, let's perform the calculation for a light intensity of 1000 lumen in the three bulbs.

consumed potence

         incandescent 100W = fluorescent (CFL) 20W = Led 12 W

Now if we can solve your exercise

a) Let's find the power consumed by each type of bulb in the 25000 h

          E = P t

incandescent

             E_incandescent = 100 25000

             E_incandescent = 2.5 105 Wh

             E_incandescent = 2.5 103 kWh

Fluorescent (Low Power) CFL

               E_CFL = 20 25000

               E _DFL = 5 102 kWh

LED

               E_LED = 12 25000

               E_LED = 3 102 kwh

               

Let's use a direct rule of proportions (rule of three) for the cost of energy, if 1 kWh costs $ 0.15, the energy calculates how much it costs

incandescent

                 Cost_Incandescent = 2.5 103 kW / h ($ 0.15 / 1kWh)

                 Cost_incandescent = $ 3.75 102 = $ 375

CFL

                  Cost_CFL = 5 102 0.15

                  Cost _CFL = $ 75

LED

                  Cost_LED = 3 102 0.15

                   Cost_LED = $ 45

Here we do not take into account the possibility of burning incandescent bulbs and there is an extra replacement cost

b) for this part we use direct proportion rules

If the day has 24 hours and the light is on for 5%, how many days are 25000 hours?

     for 5% = 0.05

hours on in a day

         #_hoursdays = 24 hrs 0.05 = 1.2 hrs

total days

          total = 25000 / 1.2

          total = 20833 days

for 30% = 0.30

       # _hoursday = 24 0.30 = 7.2 h

       Total_days = 25000 h (1 day / 7.2 h)

        total_day = 3472 days

for 60% = 0.602

         #_hours = 24 0.60 = 14.4 h

         Total _days = 25000h (1day / 14.4 h)

          Total_days = 1736 d

c) you should decide between the CFL and the Led are much lower.

Between these two should take into account the cost of the bulbs, if we only take into account the cost of the energy consumed, the selection should be LED

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A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m
kifflom [539]

<u>Answer</u>:

(a) magnitude of F = 797 N

(b)the total work done  W = 0

(c)work done by the gravitational force =  -1.55 kJ

(d)the work done by the pull  = 0

(e) work your force F does on the crate = 1.55 kJ

<u>Explanation</u>:

<u>Given</u>:

Mass of the crate, m =  220 kg

Length of the rope, L = 14.0m

Distance, d =  4.00m

<u>(a) What is the magnitude of F when the crate is in this final position</u>

Let us first determine vertical angle as follows

=>Sin \theta = \frac{d }{L}

=> \theta = Sin^{-1} \frac{d}{L} =

Now substituting thje values

=> \theta = Sin^{-1} \frac{4}{12} =

=> \theta = Sin^{-1} \frac{1}{3}

=> \theta = Sin^{-1}(0.333)

=> \theta = 19.5^{\circ}

Now the tension in the string resolve into components

The vertical component supports the weight

=>Tcos\theta = mg

=>T = \frac{mg}{cos\theta}

=>T = \frac{230 \times 9.8 }{cos(19.5)}

=>T = \frac{2254 }{cos(19.5)}

=>T = \frac{2254 }{0.9426}

=>T =2391N

Therefore the horizontal force

F = TSin(19.5)

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

<u>c) The work done by the gravitational force on the crate</u>

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values                                                            

= -230 \times 9.8\times 12 ( 1 - cos(19.5) )

= -230 \times 9.8\times 12 ( 1 - 0.9426) )

= -230 \times 9.8\times 12 (0.0574)

= -230 \times 9.8\times 0.6888

=  -230 \times 6.750

= -1552.55 J

The work done by gravity = -1.55 kJ

<u>d) the work done by the pull on the crate from the rope</u>

Since the pull  is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg  

WF = -(-1.55)

WF = 1.55 kJ

<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>

Here the work done by force is not equal to F*d  

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)      

7 0
3 years ago
Two teams are playing tug of war. The team on the right side is pulling with a force of 4332 N. The team on the left is pulling
OleMash [197]
As the greater force of tension (by 81N) is exerted by the team on the right the rope will move to the right.
6 0
3 years ago
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A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

6 0
3 years ago
Balok diam diatas bidang miring pada sudut kemiringan 40° balok mulai bergerak,tentukan koefisien gesek statis antara balok dan
Pachacha [2.7K]

Answer:

0.84

Explanation:

m = Massa balok

g = Percepatan gravitasi

\theta = Sudut kemiringan

\mu = Koefisien gesekan statik antara balok dan bidang miring

Gaya balok karena beratnya diberikan oleh

F=mg\sin\theta

Gaya gesekan diberikan oleh

f=\mu mg\cos\theta

Kondisi dimana balok mulai bergerak adalah ketika gaya balok akibat beratnya sama dengan gaya gesek pada balok.

mg\sin\theta=\mu mg\cos\theta\\\Rightarrow \mu=\dfrac{mg\sin\theta}{mg\cos\theta}\\\Rightarrow \mu=\tan\theta\\\Rightarrow \mu=\tan40^{\circ}\\\Rightarrow \mu=0.84

Koefisien gesekan statik antara balok dan bidang miring adalah 0.84.

7 0
3 years ago
Who was the discoverer of the neutron
Pavel [41]

Answer:

James Chadwick

Explanation:

In May 1932 James Chadwick announced that the core also contained a new uncharged particle, which he called the neutron

6 0
2 years ago
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