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Mice21 [21]
2 years ago
14

An 11.0 -W energy-efficient fluorescent lightbulb is designed to produce the same illumination as a conventional 40.0-W incandes

cent lightbulb. Assuming a cost of 0.110 kWh for energy from the electric company, how much money does the user of the energy-efficient bulb save during 100h of use?
Physics
1 answer:
yarga [219]2 years ago
6 0

The amount or cost that the user of the energy-efficient bulb save during 100h of use will be $0.319.

<h3>How to calculate the cost?</h3>

For the 11.0W bulb, it should be noted that the value will be:

= 11.0 × 100 × (1/1000) × 0.110

= $0.121

The 40W bulb will be:

= 40 × 100 × (1/1000) × 0.110

= $0.44

Therefore, the amount that will be saved will be:

= $0.44 - $0.121

= $0.319

Learn more about cost on:

brainly.com/question/25109150

#SPJ4

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Newton’s third law is a force is a push or pull that acts upon an object as a result of another object.. A shot gun when fired pulls back when a force (you pulling the trigger) is acted upon it. This is how it relates. A rifle has less kick than a shot gun because the rifle is smaller and has less of a force than a shot gun.
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3 years ago
Which statement best describes metallic bonding?
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THE ANSWER IS D. A NONMETAL ATOM TRANSFERS ELCTRONS TO A METAL ATOM

ITS A TYPE OF ENERGY BONDING THAT ARISES FROM ELECTROSTATIC. ATTRACTIVE FORCE BETWEEN CONDUCTION ELCTRONS (A NONMETAL ATOM) AND A POSITIVELY CHARGED METAL ATOM ( WHO ATTRACTS THE NONMETALS ELECTRONS BECAUSE ITS POSITIVELY CHARGED)

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3 years ago
Read 2 more answers
1. How much energy would be required to melt 450 grams of ice at 0°C?
xenn [34]

Answer:

Explanation:

1. The amount of heat needed to melt ice at 0°C is equal to the mass of the ice times the latent heat of fusion.

q = mL

q = (450 g) (334 J/g)

q = 150,300 J

q = 150 kJ

2. The amount of heat released by the condensation of steam at 100°C is equal to the mass of the steam times the latent heat of vaporization.

q = mL

q = (325 g) (2260 J/g)

q = 734,500 J

q = 735 kJ

3. q = mL

q = (85 g) (2260 J/g)

q = 192,100 J

q = 190 kJ

4. q = mL

q = (225 g) (334 J/g)

q = 75,150 J

q = 75.2 kJ

5. Above 100°C, water is steam.  The amount of heat needed to increase the temperature of steam is equal to its mass times its specific heat times the change in temperature.

q = mCΔT

q = (20.0 g) (2.03 J/g/°C) (303.0°C − 283.0°C)

q = 812 J

6. q = mCΔT

q = (15.0 g) (2.03 J/g/°C) (250.0°C − 275.0°C)

q = -761 J

7. q = mCΔT

q = (10.0 g) (0.90 J/g/°C) (55°C − 22°C)

q = 297 J

8. q = mCΔT

198 J = (55.0 g) C (15°C)

C = 0.24 J/g/°C

9. q = mCΔT

41,840 J = m (4.184 J/g/°C) (28.5°C − 22.0°C)

m = 1540 g

10. q = mCΔT

q = (193 g) (2.46 J/g/°C) (35°C − 19°C)

q = 7600 J

11. First, the temperature of the ice must be raised to 0°C.

q = mCΔT

q = m (2.09 J/g/°C) (0°C − (-23.0°C))

q/m = 48.1 J/g

Next, the ice must be melted.

q = mL

q/m = 334 J/g

Then, the water must be heated to 100°C.

q = mCΔT

q = m (4.184 J/g/°C) (100°C − 0°C)

q/m = 418.4 J/g

The water is then vaporized.

q = mL

q/m = 2260 J/g

Finally, the steam is heated to its final temperature.

q = mCΔT

q = m (2.03 J/g/°C) (118°C − 100°C)

q/m = 36.5 J/g

So the total amount of energy needed is:

q/m = 48.1 J/g + 334 J/g + 418.4 J/g + 2260 J/g + 36.5 J/g

q/m = 3100 J/g

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3 years ago
A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i
Sunny_sXe [5.5K]

Answer:

114.44 J

Explanation:

From Hook's Law,

F = ke................. Equation 1

Where F = Force required to stretch the spring, k = spring constant, e = extension.

make k the subject of the equation

k = F/e.............. Equation 2

Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.

Substitute into equation 2

k = 44.5/1.016

k = 43.799 N/m

Work done in stretching the 9 in beyond its natural length

W = 1/2ke²................. Equation 3

Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m

Substitute into equation 3

W = 1/2×43.799×2.286²

W = 114.44 J

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Why does a lunar eclipse last for a few hours whereas a solar eclipse only lasts for a few minutes?
Aneli [31]

Answer:

The type and length of a lunar eclipse depend on the Moon's proximity to either node of its orbit. ... A total lunar eclipse can last up to nearly 2 hours, while a total solar eclipse lasts only up to a few minutes at any given place, due to the smaller size of the Moon's shadow.

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