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Anarel [89]
2 years ago
5

You have a Switch LED Lightbulb with the following information: It is equivalent to a 40 Watt incandescent lightbulb. It uses 8

Watts of electricity. It is estimated to last 25,000 hours. It is dimmable. It produces 450 lumens of light. It has a color temperature of 2,700K (warm white). Based on this information, calculate the lighting efficiency of this bulb and determine whether this is more or less efficient than a traditional incandescent lightbulb.
56 lumens/Watt, less efficient

56 lumens/Watt, more efficient

0.0178 lumens/Watt, less efficient

0.0178 lumens/Watt, more efficient​
Engineering
1 answer:
DochEvi [55]2 years ago
4 0

The lightning efficiency based on the scenario depicted will be C. 56 lumens/Watt, more efficient.

<h3>How to calculate the lightning efficiency</h3>

The efficiency of the incandescent bulb will be:

= 450/40 = 11.25 lumens per watt.

The efficiency of the LED bulb will be:

= 450/8 = 56 lumens per watt.

In this case, the LED bulb is more efficient than the incandescent bulb.

Therefore, the lighting efficiency will be 56 lumens/Watt, more efficient

Learn more lightning efficiency on:

brainly.com/question/25927632

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A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
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Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

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