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FromTheMoon [43]
3 years ago
10

some fuel cells are powered by hydrogen. scientists are looking into the decomposition of water (h2o) to make hydrogen fuel with

the help of the sun’s energy. what challenge do scientists face with this task?
Physics
2 answers:
sesenic [268]3 years ago
4 0

Answer : First of all, to harness the solar energy and utilize it for the decomposition of water to make hydrogen fuel cell seems to be easy but practically it has many hurdles in its way. It might not work well, when the days would not be sunny. Also, the production of hydrogen fuel would require decomposing water which may need a high energy of activation, which would be more than the energy that will be produced from the fuel cell.

Bogdan [553]3 years ago
3 0
The challenge that scientists will face with this task is the production of hydrogen since hydrolysis or the <span>decomposition of water (H2O) requires a certain amount of energy to be supplied. Scientists should look at how much is this energy supplied. It can be more than what would be produced from the fuel cells which will eventually result to a loss.</span>
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Why can light microscopes produce images in their natural color
Fiesta28 [93]

Answer:

The electron microscopes  are uses the electron to produce the image of the any object.The area where the electron passes then that area  appeaser white but on the other hand where the electron does not pass that area appears black.But this does not give any information about the color of light it shows only white and black areas.We also know that color is the property of photons ( light).So the microscopes does not produce color of the image.

6 0
3 years ago
(a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of en
inysia [295]

Answer:

4.42 x 10⁷ W/m²

Explanation:

A = energy absorbed = 500 J

η = efficiency = 0.90

E = Total energy

Total energy is given as

E = A/η

E = 500/0.90

E = 555.55 J

t = time = 4.00 s

Power of the beam is given as

P = E /t

P = 555.55/4.00

P = 138.88 Watt

d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m

Area of the circular spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Intensity of the beam is given as

I = P /A

I = 138.88 / (3.14 x 10⁻⁶)

I = 4.42 x 10⁷ W/m²

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