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jok3333 [9.3K]
3 years ago
15

An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 100-W incandescent bulb can be replac

ed by a 16-W LED bulb. Both produce 1600 lumens of light. Assuming the cost of electricity is $0.10 per kilowatt-hour, how much does it cost to run the bulb for one year if it runs for four hours a day?
Physics
1 answer:
Gelneren [198K]3 years ago
6 0

Answer:

Explanation:

convert 16W to kilowatt

16 W / 1000 = 0.016 kW

It runs for 4 hours a day and for one year

so for 1 year, it runs for = 4 × 365 = 1460 hr

 E ( kilowatt-hr ) = power × time = 0.016 kw × 1460 = 23.36 kwhr

$ 0.10 = 1 kilowatt-hr

the amount it will cost for a year = 23.36 kwhr  × $ 0.10 = $ 2.336

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You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If
wlad13 [49]

Answer:

  • between locations that are 14 cm outboard of the chair edges
  • the weightless board is centered and end sitters are 25 cm from the ends

Explanation:

We can assume the .5 m-wide chair means that it is comfortable for each student to sit 0.25 m from the end of the board. If the board is centered on the chair, then each student is 1 m from the edge of the chair.

When Dan and Tahreen are seated on the board, their center of mass is ...

  (50 kg×2.5 m)/(50 kg +67 kt) = 1.068 m

to the right of the position where Dan is seated. Since this location is over the chair, the board is stable.

Komila can sit as much as x distance from the chair toward Dan, where ...

  67(1) +54(x) = 50(1.5)

  x = 8/54 ≈ 0.148 . . . . meters

Or, Komila can sit as much as x distance from the chair toward Tahreen, where ...

  67(1.5) = 54(x) +50(1)

  x = 50.5/54 ≈ 0.935 . . . . meters

<u>Scenario 1</u>

Assuming the (weightless) board is centered on the chair, Komila can sit anywhere between 14.8 cm left of the chair and 93.5 cm right of the chair and the board will remain stable. Sitting on the board centered on the chair is a suitable location. The two students sitting on the ends must become (and stay) seated at the same time. They both must be seated 0.25 m from the end of the board for the other dimensions to remain valid.

<u>Scenario 2</u>

Assuming the (weightless) board is located so its left end is 1.068 m from the chair, and Dan and Tahreen are seated 0.25 m from the ends of the board, Komila can sit anywhere within (117/54×.25 m) = 0.54 m of the chair and the board will remain stable. Again, sitting centered on the chair is a suitable location.

__

There does not appear to be any location where Komila can sit and have the board remain stable with only Dan or Tahreen seated on one end (assuming a width of 0.5 m for each sitter).

_____

<em>Comment on the question</em>

For the board to remain stable, the sum of moments about either edge of the chair must tend to rotate the board toward the chair. This sum will depend on the locations of the sitters relative to each edge of the chair, so there is significant freedom in choosing locations. To make the problem tractable, we have made some specific assumptions about where the board is and what the locations of the sitters might be. YMMV

3 0
4 years ago
Newton's "quantity of motion" is conservation.<br><br> True<br><br> False
Valentin [98]
The answer is false hope this helps
6 0
3 years ago
Read 2 more answers
A 8.0\,\text {kg}8.0kg8, point, 0, start text, k, g, end text box is released from rest at a height y_0 =0.25\,\text my 0 ​ =0.2
Ratling [72]

Answer:

μ = 0.125

Explanation:

To solve this problem, which is generally asked for the coefficient of friction, we will use the conservation of energy.

Let's start working on the ramp

starting point. Highest point of the ramp

         Em₀ = U = m h y

final point. Lower part of the ramp, before entering the rough surface

        Em_{f} = K = ½ m v²

as they indicate that there is no friction on the ramp

          Em₀ = Em_{f}

          m g y = ½ m v²

          v = \sqrt{2gy}

we calculate

          v = √(2 9.8 0.25)

           v = 2.21 m / s

in the rough part we use the relationship between work and kinetic energy

          W = ΔK = K_{f} -K₀

as it stops the final kinetic energy is zero

          W = -K₀

The work is done by the friction force, which opposes the movement

          W = - fr x

friction force has the expression

          fr = μ N

let's write Newton's second law for the vertical axis

         N-W = 0

         N = W = m g

we substitute

            -μ m g x = - ½ m v²

           μ = \frac{v^{2} }{2 g x}

Let's calculate

           μ = \frac{2.21^{2}}{2\  9.8\  2.0}

           μ = 0.125

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Does Newton’s 1st Law of Motion apply to sports
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Answer: Yes indeed it does.

Explanation: Newton's Three Laws of Motion explain how forces create motion in sport. These laws are usually referred to as the Laws of Inertia, Acceleration, and Reaction. maybe brainliest to level up.

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Why do organisms need to breathe?​
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Answer:

Most living things need oxygen helps organisms grow,reproduce,and turn food into energy.Oxygen gives our cells the ability to break down food in order to get the energy we need to survive.

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