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3 years ago

Infrared telescopes are usually placed on high-flying airplanes or on satellites in space __________

Physics

1 answer:

Yuliya22 [10]3 years ago

3 0

Answer:

In order to improve visibility

Explanation:

Infrared telescopes are made using infrared cameras that contain infrared detectors which are solid-state and are maintained at very cold (cryogenic) temperatures

Infrared radiation is absorbed by water vapor which is present in the Earth's atmosphere, leading to the limitation of the use of infra red telescopes at high altitudes such as mountains, high flying planes or satellites

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Which scenario presents a possible environmental impact due hydrogen fuel cells?

Mekhanik [1.2K]

The scenario that presents a possible environmental impact due to hydrogen fuel cells is that, the electricity used to supply the fuel cells with Hydrogen is produced using fossil fuels.

The question is incomplete, but these should have been the options;

a.) The fuel cells are used to power internal combustion engines rather than electric motors.

b.) The electricity used to supply the fuel cells with hydrogen is produced using fossil fuels.

c.) The combined exhaust emissions from fleets of buses equipped with fuel cells generates smog.

d.) None of these are correct.

The answer therefore is option B. The Hydrogen fuel cell is a device that uses Hydrogen as an energy source to power machines and to generate heat and electricity. It is a green energy alternative to fossil fuels as unlike fossil fuels, it doesn't produce green house gases and is non-toxic. Also, unlike other clean sources of energy like wind and water, it doesn't require a large area of land to be produced.

Hydrogen fuel cells work by combining hydrogen and oxygen atoms. The resulting energy produced is converted to electrical energy which can be used to power electrical motors, while the by-product, water is non-toxic, renewable and doesn't create smog.

However, to produce the hydrogen used in the cells, high amount of energy is required in order to separate hydrogen from oxygen. To accomplish this, fossil fuels such as coal, natural gas and oil are used, releasing harmful emissions which creates a negative environmental impact. This makes the answer to the above question option B. A better alternative though, is to use other renewable energy sources like wind or solar.

To learn more about the benefits of Hydrogen fuel cells, visit;

brainly.com/question/6501095

#SPJ4

8 0

1 year ago

10. Solve the following numerical problems

frosja888 [35]

Answer:

$\boxed {\boxed {\sf 120 \ Joules}}$

Explanation:

Work is equal to the product of force and distance.

$W=F*d$

The force is 8 Newtons and the distance is 15 meters.

$F= 8 \ N \\d= 15 \ m$

Substitute the values into the formula.

$W= 8 \ N * 15 \ m$

Multiply.

$W= 120 \ N*m$

1 Newton meter is equal to 1 Joule
Our answer of 120 N*m equals 120 J

$W= 120 \ J$

The work done is <u>120 Joules</u>

3 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$