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3 years ago

What technologies are astronomers using to look for life on other planets? Why? help!!! please

Physics

1 answer:

Brilliant_brown [7]3 years ago

6 0

<h3>Technology used by astronomers to look for life on other plants:</h3>

Space-based Hubble and Spitzer telescopes and Green Bank telescopes are used.
Many new satellites are launched to trace earth like planets. Even the telescopes are carried by spacecrafts.

Astronomers are using Allen Telescope Array that comprises 42 radio antennas to get hold of signals over a wide range of radio frequencies that are tuned to listen to the regions nearer to the red dwarf star constellations.
Scientists needed telescopes that are capable of detecting planets like Earth in our planet’s neighborhood and see if any biological signatures connected to any microbial or other intelligent life can be found.
Radio signals are also being used to detect if the outside world has something to communicate so that they can be detected.

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Every gene is made of two a).genotypes b).alleles c).cells

igomit [66]

Answer:

Every gene are made of 2 alleles

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2 years ago

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The change in momentum of an object is equal to the ____________ that acts on it.

meriva

Answer : The change in momentum of an object is equal to the impulse that acts on it.

Explanation :

Change in momentum : The change in momentum of an object is the product of the mass and the change in velocity of an object.

The formula of change in momentum is,

$\Delta p=m\times \Delta v$

Impulse : An impulse of an object is the product of the force applied on an object and the change in time. Impulse is also equivalent to the change in momentum of an object.

$J=F\times \Delta t$

Proof :

$J=F\times \Delta t\\\\J=(m\times a)\times \Delta t\\\\J=m\times (a\times \Delta t)\\\\J=m\times \Delta v=\Delta p$

Hence, the change in momentum of an object is equal to the impulse that acts on it.

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3 years ago

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The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply

kykrilka [37]

Answer: Got It!

Explanation: let s = speed at launch

v = 0 at top = s sin 63 - g t

so at top

t = s sin 63/g = .0909 s

h = 13.6 = s sin 63 t - 4.9 t^2

13.6 = .081s^2 - .0405 s^2

s^2 = 336

s = 18.3 m/s

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4 0

3 years ago

What is the function of the bronchioles? please help

beks73 [17]

Answer:

The bronchioles function is to deliver air to tiny sacs called alveoli where oxygen and carbon dioxide are exchanged.

Explanation:

Bronchioles are air passages inside the lungs that branch off like tree limbs from the bronchi—the two main air passages into which air flows from the trachea (windpipe) after being inhaled through the nose or mouth. The bronchioles deliver air to tiny sacs called alveoli where oxygen and carbon dioxide are exchanged.

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2 years ago

A sound is recorded at 19 decibels. What is the intensity of the sound?

sp2606 [1]

$1 \times 10^{-10.1} \mathrm{Wm}^{-2}$ is the intensity of the sound.

Answer: Option B

Explanation:

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about $1 \times 10^{-12} \mathrm{Wm}^{-2}$ ). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

$\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)$

Where,

I = Intensity of the sound produced

$I_{0}$ = Standard Intensity of sound of 60 decibels = $1 \times 10^{-12} \mathrm{Wm}^{-2}$

So for 19 decibels, determine I as follows,

$19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9$

When log goes to other side, express in 10 to the power of that side value,

$\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}$

$I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}$

5 0

3 years ago