All categories

3 years ago

Suppose the price of electricity is 12.75 cents per kWºh. What

Physics

1 answer:

adell [148]3 years ago

5 0

Answer:

The difference between the cost of operating LED and incandescent bulb is $5.1

Explanation:

We are given the cost of electricity that is 12.75 cents per kWh. We want to find out the difference in the operating cost of an incandescent and LED bulb for a time period of 2,000 hours.

Since we are not given the rating of the incandescent bulb and LED bulb, we will assume their ratings.

For a light intensity of 250 Lumens;

The average rating of an LED bulb is approximately 5 Watts.

The average rating of an incandescent bulb is approximately 25 Watts.

Now lets find out the kWh of each bulb.

Energy = Power×Time

For LED bulb:

E = 5×2,000 = 10,000 Wh

Divide by 1000 to convert into kWh

E = 10,000/1000 = 10 kWh

Cost = 12.75×10 = 127.5 cents

Cost = $1.27

For Incandescent bulb:

E = 25×2,000 = 50,000 Wh

Divide by 1000 to convert into kWh

E = 50,000/1000 = 50 kWh

Cost = 12.75×50 = 637.5 cents

Cost = $6.37

Difference in Cost:

Difference = $6.37 - $1.27 = $5.1

Therefore, the difference between the cost of operating LED and incandescent bulb is $5.1.

You might be interested in

Im bad at work problems can any one help with this problem ?

lyudmila [28]

We will put the number of trips in the first column, the miles driven in the second column and gallons of fuel used in the third column.

8 7,680 1,010
7 9,940 1,330
12 14,640 1,790
12 13,920 2,050

4 0

3 years ago

Which claim would most likely be considered valid?

Mekhanik [1.2K]

Please state the options and I will answer to the best of my abilities XD

7 0

3 years ago

Read 2 more answers

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago

Three substances that can make electricity. What are these substance

DaniilM [7]

Answer:

Ionic substances, also known as salts, conduct electricity when dissolved in water or when in the liquid state

Explanation:

7 0

3 years ago

If an object is placed between the focal point and twice the focal length of a convex lens, which type of image will be produced

valentinak56 [21]

<span>The image produced is real and enlarged.

Check for various positions of objects and Images for convex lens.

Note at position of 2F, the image is same as the object, and once it is between 2F and F, the image becomes bigger than the object. </span>

3 0

3 years ago

Read 2 more answers