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Montano1993 [528]
3 years ago
15

5. Why might it be useful to determine the elements that a planet or moon is made up of?

Physics
2 answers:
jekas [21]3 years ago
6 0

One of the main interests in science relating to space, is finding other planets that are similar to Earth (similar atmosphere and can sustain water and in turn the potential of life)

So if we are able to determine composition of the planet, we might be able to see if it can or does have water on it - which could mean life can survive on the planet. Which opens up options of populating another planet, or discovering a new life-form/ living organism.

exis [7]3 years ago
3 0

because we will know more about out solar system

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When do sunspots disappear?
dedylja [7]

Answer:

In 5 years or so, the sun will be awash in sunspots and more prone to violent bursts of magnetic activity.

Explanation

once the magnetic field weakens the area and cold plasma enters the area of the sunspot

8 0
2 years ago
an object is moving with initial velocity of 5 m/s. After 10 seconds final velocity is 10 m/s. Calculate its acceleration.​
Brut [27]

Answer:

0.5m/s2

Explanation:

acceleration= change in velocity/time taken

= v - u/ t

= 10-5/10

=5/10

= 0.5m/s2

6 0
3 years ago
What are the top two gasses in the earth’s atmosphere ?
ch4aika [34]

Answer:

Nitrogen and oxygen are by far the most common; dry air is composed of about 78% nitrogen (N2) and about 21% oxygen (O2). Argon, carbon dioxide (CO2), and many other gases are also present in much lower amounts; each makes up less than 1% of the atmosphere's mixture of gases.

7 0
3 years ago
Sunlight strikes a solar panel, which is then used to power the lights in a
Stells [14]

Answer:

B

Explanation:

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7 0
2 years ago
Read 2 more answers
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

3 0
2 years ago
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