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Sidana [21]
4 years ago
7

Zeros between a decimal point and a non zero number are _______ significant

Physics
1 answer:
KiRa [710]4 years ago
6 0
They are indeed significant. This statement is true. The zero that is not significant is the one found before the first non-zero digit.
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the orion nebula (at least the part we can see) is not very old (yet). while several hot, massive stars have had a chance to for
svetoff [14.1K]

One of the brightest nebulae in the night sky, the Orion Nebula may be seen with the unaided eye. The Trapezium is a young open cluster of four main stars in this magnitude 4 interstellar cloud of ionized atomic hydrogen.

<h3>What is the source of the Orion Nebula's crimson glow?</h3>
  • The hydrogen gas in the Orion Nebula, which is powered by radiation from young stars, gives off a crimson tint. The nebula's blue-violet regions are reflecting radiation from bright, blue-white O-type stars while the red areas are emitting light.
  • The Orion Nebula is one of many massive clouds of gas and dust in our Milky Way galaxy, say contemporary astronomers, and is one of the largest. It is approximately 1,300 light years away from Earth. This enormous hazy cocoon, which measures approximately 30 to 40 light-years in diameter, is generating potentially a thousand stars.  

To learn more about Orion nebula refer to:

brainly.com/question/15575332

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3 0
1 year ago
Fill-in-the-blank test questions measure ________; matching concepts with their definitions measures ________.
Ulleksa [173]

Test questions measure recall; matching concepts with their definitions measures recognition.

<u>Explanation: </u>

According to Psychology our brain remembers everything what we learn but the understanding and remembering the right answer for the right question needs training and understanding ability. So in order to enhance the ability of recalling and recognizing among the students, the concept of test questions and matching with definitions are used in curricular activities.

As the students will be learning different terms, definitions, methods and different subjects, they should be able to distinguish among different definitions as well as they should recall the things they have learnt. So the answers for the test questions will help to recall the topics learnt by the students while the matching concept will help the students to recognize each definition with their terms.

6 0
3 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
Sound waves can travel through which mediums?
aksik [14]

Answer:

B. Solids, liquids, and gases.

Explanation:

I have no explanation.

5 0
3 years ago
Find the work done lifting a 290-pound weight 28 feet in the air.
Nataly [62]
The answer is d.8,120 foot-pounds
6 0
3 years ago
Read 2 more answers
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