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Simora [160]
3 years ago
12

Problem 6.13 The point of this question is to compare rest energy and kinetic energy at high speeds. An alpha particle (a helium

nucleus) is moving at a speed of 0.9998 times the speed of light. Its mass is 6.40 × 10-27 kg. (a) What is its rest energy?
Physics
1 answer:
Kryger [21]3 years ago
3 0

Answer: E = 5.76×10^-10 J

Explanation: rest energy formula is given below as

E = mc²

m = mass of object = 6.40×10^-27 kg

c = speed of light = 3×10^8 m/s

E = 6.40×10^-27 × (3×10^8)²

E = 6.40×10^-27 × 9×10^16

E = 57.6 ×10^-11

E = 5.76×10^-10 J

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When astronomers look at distant galaxies, what sort of motion do they see?
arlik [135]
Hello! You can call me Emac or Eric.

I understand your problem, that question is pretty hard. But I found some information that I think you should read. This can get your problem done quickly.

Please hit that thank you button if that helped, I don’t want thank you’s I just want to know that this helped.

Please reply if this doesn’t help, I will try my best to gather more information or a answer.

Here is some good information that could help you out a lot!


Let’s begin by exploring some techniques astronomers use to study how galaxies are born and change over cosmic time. Suppose you wanted to understand how adult humans got to be the way they are. If you were very dedicated and patient, you could actually observe a sample of babies from birth, following them through childhood, adolescence, and into adulthood, and making basic measurements such as their heights, weights, and the proportional sizes of different parts of their bodies to understand how they change over time.

Unfortunately, we have no such possibility for understanding how galaxies grow and change over time: in a human lifetime—or even over the entire history of human civilization—individual galaxies change hardly at all. We need other tools than just patiently observing single galaxies in order to study and understand those long, slow changes.

We do, however, have one remarkable asset in studying galactic evolution. As we have seen, the universe itself is a kind of time machine that permits us to observe remote galaxies as they were long ago. For the closest galaxies, like the Andromeda galaxy, the time the light takes to reach us is on the order of a few hundred thousand to a few million years. Typically not much changes over times that short—individual stars in the galaxy may be born or die, but the overall structure and appearance of the galaxy will remain the same. But we have observed galaxies so far away that we are seeing them as they were when the light left them more than 10 billion years ago.


That is some information, I do have more if you need some! Thanks!

Have a great rest of your day/night! :)


Emacathy,
Brainly Team.


8 0
3 years ago
Help me please I can't get the final step​
inna [77]

Answer:

\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

\displaystyle A=\frac{3}{2}B^mC^n

and the dimensions of each variable is:

A=L^2T^2

B=LT^{-1}

C=LT^2

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n

Operating:

L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)

L^2T^2=L^{m+m}T^{-m+2n}

Equating the exponents:

m+n=2

-m+2n=2

Adding both equations:

3n=4

Solving:

n=4/3

m=2-4/3=2/3

Answer:

\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}

6 0
3 years ago
A6 kg mass moving at 10m/s collides with a 4 kg mass moving in the
Nat2105 [25]

Answer:

Explanation:

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4 0
3 years ago
After a hold up, Robin Banks flees in his 1575 kg getaway car at 20 m/s. He crashes into a 45 kg highway barrel which is at rest
fiasKO [112]

Answer:

v = 38.9 m/s

Explanation:

7 0
3 years ago
Read 2 more answers
A 20-kg child running at 1.4 m/s jumps onto a playground merry-go-round that has mass 180 kg and radius 1.6m. She is moving tang
Dominik [7]

Answer:

ωf = 0.16 rad/s

Explanation:

Moment of inertia of the child = mr² = 20(1.6²) = 51.2 kg•m²

Moment of Inertia of the MGR = ½mr² = ½(180)1.6² = 230.4 kg•m²

(ASSUMING it is a uniform disk)

Initial angular momentum of the child = Iω = I(v/r) = 51.2(1.4/1.6) = 44.8 kg•m²/s

Conservation of angular momentum

44.8 = (51.2 + 230.4)ωf

ωf = 0.15909090...

4 0
3 years ago
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