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SVETLANKA909090 [29]
3 years ago
6

A starlike object seen on deep sky photographs coincides with an intense radio source and has a spectrum in which the characteri

stic lyman pattern of hydrogen spectral lines has been shifted from the ultraviolet to the visible spectral range. what is this object?
Physics
1 answer:
liberstina [14]3 years ago
6 0
A quasar - <span>Quasars are very distant objects. They  are really bright masses of energy and light. The word quasar is actually short for quasi-stellar radio source or quasi-stellar object. This high amount of energy in such a small area can only be formed by matter falling towards a Black Hole. That's why quasars are believed to be huge black holes  in the center of young galaxies.
</span><span />
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How many atoms of carbon would two molecules of glucose (C6H12O6) have?
Sauron [17]
Well according to the molecular formula of glucose, one molecule would have 6 carbon atoms, and thus 2 molecules of glucose would have 12 carbon atoms.

The correct response would be B. 12.
4 0
3 years ago
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A car traveling at 100 km/hr strikes and unfortunate bug and splatters it.the net force of impact is
Paladinen [302]

Answer:

it's zero

Explanation:

it is there is your answer

7 0
2 years ago
Water being turned into ice cubes in a freezer is an example of _____.
Gekata [30.6K]

Answer:

a physical change

Explanation:

after the water turns to ice, it will melt and became water again making which means it's reversible this being. a physical change

7 0
2 years ago
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How is it possible to get three precise but inaccurate measurements of the same volume of water
Lubov Fominskaja [6]
The inaccurate measurements must be similar to the other two measurements (ex; 590, 589, 599), but different from the actual volume of water. (Ex; the actual volume is let say.. 100, but you measured 50, 49, 40)
8 0
3 years ago
Guys please help me ​
Likurg_2 [28]

Answer:

1)t=2.26\: s

2)S=33.9\: m

3)v=26.77\: m/s

4)\alpha=55.92

Explanation:

1)

We can use the following equation:

y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

0=25-0.5*9.81*t^{2}

t=2.26\: s

2)

The equation of the motion in the x-direction is:

v_{ix}=\frac{S}{t}

15=\frac{S}{2.26}

S=33.9\: m

3)

The velocity in the y-direction of the stone will be:

v_{fy}=v_{iy}-gt

v_{fy}=0-(9.81*2.26)

v_{fy}=-22.17\: m/s

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}

v=26.77\: m/s

4)

The angle of this velocity is:

tan(\alpha)=\frac{22.17}{15}

\alpha=tan^{-1}(\frac{22.17}{15})

\alpha=55.92

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0
3 years ago
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