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borishaifa [10]
3 years ago
7

If scientists could create a sheet of glass with a refractive index so high that it took light months or years to travel through

the glass, what kind of things do you think they could do with it?
Physics
1 answer:
omeli [17]3 years ago
4 0

Wow ! They could set up sheets of "slow glass" beside beautiful forests with rivers and squirrels and deer and grassy fields, and load a year of this scene into the glass, and then sell it to people who live next to dirty brick walls or ugly empty lots, and those people could install the slow glass in their windows and have beautiful scenery, until it all worked its way out of the glass.

This is a great idea ! If you possibly can, find and read the sci-fi short story "Light of Other Days" written by Bob Shaw, published in Analog Science Fiction in 1966. It's all about this exact type of glass. I read this story in 1966 and I never forgot it ! (Not yet anyway.)

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Answer:

1,050 Joules

Explanation:

<u>Step 1:</u> work done in moving the box 30 meters

work done = force X distance

                  = 25N X 30 = 750 Joules

<u>Step 2: </u>calculate total internal energy

Total internal energy = work done + kinetic energy

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3 years ago
True or false, only a radioactive isotope will have a half-life
maks197457 [2]
False not radioactive isotope will have a half-life
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3 years ago
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar
defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

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\zeta = \dfrac{W_{out}}{Q_{supplied }}

Q_{supplied } = \dfrac{2}{0.03} \ MW

Q_{supplied } = 66.66 \ MW

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

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LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}

Also;

\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K

\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K

LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}

LMTD = \dfrac{8}{In (5)}

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Thus, the required heat exchanger area A is calculated by using the formula:

Q_H = UA (LMTD)

where;

U = overall heat coefficient given as 1200 W/m².K

66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\  A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\  \mathbf{A = 11178.236 \ m^2}

The mass flow rate:

Q_{H} = mC_p(T_{in} -T_{out} )  \\ \\  66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{  66.667 \times 10^6}{4.18 \times 8} \\ \\  \mathbf{m = 1993630.383 \ kg/s}

3 0
3 years ago
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vector is the answer of this blank

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