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3 years ago

How can spectroscopy and infrared technology be useful in space?

Physics

1 answer:

olchik [2.2K]3 years ago

3 0

Answer:

Explanation:

Spectroscopy measures electromagnetic radiation.

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HEEEEEEELLLLLLLPPPPP

Firdavs [7]

Answer:

the less shielding of electrons

6 0

2 years ago

What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)

Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity $g=10m/sec^2$

And final value of acceleration due to gravity $g=9.8m/sec^2$

We have to find the percentage error

We know that percentage error is given by $percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100$

So $percentage\ error=\frac{10-9.8}{10}\times 100=2$ %

4 0

3 years ago

Need help ASAP

VARVARA [1.3K]

Answer:

the correct one is D,

Ultraviolet, x-ray, gamma ray

Explanation:

Electromagnetism radiation are waves of energy that is expressed by the Planck relationship

E = h f

where h is the plank constant and f the frequency of the radiation.

Also the speed of light is

c = λ f

we substitute

E = h c /λ

therefore to damage the cells of the body radiation of appreciable energy is needed

microwave radiation has an energy of 10⁻⁵ eV

infrared radiation E = 10⁻² eV

visible radiation E = 1 to 3 eV

radiation Uv E = 3 to 6 eV

X-ray E = 10 eV

gamma rays E = 10 5 eV

therefore we see that the high energy radiation is gamma rays, x-rays and ultraviolet light.

When checking the answers, the correct one is D

6 0

2 years ago

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

FrozenT [24]

Answer:

$\omega=0.31\frac{rad}{s}$

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

$a_c=\frac{v^2}{r}(1)$

Here, r is the radius and v is the tangential speed, which is given by:

$v=\omega r(2)$

Here $\omega$ is the angular velocity, we replace (2) in (1):

$a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r$

Recall that $r=\frac{d}{2}=\frac{200m}{2}=100m$ .

Solving for $\omega$ :

$\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}$

3 0

3 years ago

Sound waves cannot carry energy through. A water B air C a mirror D a vacuum

Ber [7]

I looked up the question and got D- a vacuum

3 0

2 years ago