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Radda [10]
3 years ago
6

Stars originate as large bodies of slowly rotating gas.Because of gravity, these clumps of gas slowly decrease in size.The angul

ar velocity of a star increases as it shrinks because of__.
A.) law of universal gravitation
B.) conservation of linear momentum
C.) conservation of energy
D.) conservation of mass
E.) conservation of angular momentum
Physics
1 answer:
Tamiku [17]3 years ago
4 0

Answer:

E.) conservation of angular momentum

Explanation:

The angular momentum is defined as:

L=r x mv

where r is the radius of the star, m is the mass and v the angular velocity.

and angular momentum is an amount that is conserved, so the angular momentum before the star is compressed must be equal to the angular momentum after the star was compressed:

r_{1} x mv_{1}=r_{2} x mv_{2}

the second radius is smaller than the first radius, since the star shrinked, the second angular velocity must be greater that the first.

In other words, the angular velicity increases as the star shrinks because of the conservation of angular momentum.

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Object A is positively charged. Object A and Object B
ozzi

Answer: object B is negatively charged, object C is positively charged and object D is also positively charged

Explanation: since unlike charges attract and like charges repel, for object A which is positively charged and B to attract B must be negatively charged and then for B which is negatively charged and C to attract C must be positively charged and for C and D to repel they have to be of thesame charge which means D is positive as well.

4 0
3 years ago
SHOW WORK
Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\

For Point b:

E=\frac{1}{2} m a^2 w^2\\\\

   =\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J

For Point C:

V_{max}= a w

        = (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}

For point D:

X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm

5 0
3 years ago
If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
sdas [7]
Data:
f_{2} = 42 Hz
n (Wave node)
V (Wave belly) 
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
f_{n} =  \frac{nV}{2L}

Wire 2 → 2º Harmonic → n = 2

f_{n} = \frac{nV}{2L}
f_{2} = \frac{2V}{2L} &#10;
2V =  f_{2} *2L
V =  \frac{ f_{2}*2L }{2}
V =  \frac{42*2L}{2}
V =  \frac{84L}{2}
V = 42L

Wire 1 → 1º Harmonic or Fundamental rope → n = 1


f_{n} = \frac{nV}{2L}
f_{1} = \frac{1V}{2L}
f_{1} =  \frac{V}{2L}

If, We have:
V = 42L
Soon:
f_{1} = \frac{V}{2L}
f_{1} = \frac{42L}{2L}
\boxed{f_{1} = 21 Hz}

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0
3 years ago
Read 2 more answers
I will mark you brainlist. How can you use a tuning fork to tune a piano?
Phoenix [80]

A tuning fork's job is to establish a single note that everybody can tune to.

Most tuning forks are made to vibrate at 440 Hz, a tone known to musicians as "concert A." To tune a piano, you would start by playing the piano's "A" key while ringing an "A" tuning fork. If the piano is out of tune, you'll hear a distinct warble between the note you're playing and the note played by the tuning fork; the further apart the warbles, the more out-of-tune the piano. By either tightening or loosening the piano's strings, you reduce the warble until it's in line with the tuning fork. Once the "A" key is in tune, you would then adjust all of the instrument's 87 other keys to match. The method is much the same for most other instruments. Whether you're tuning a clarinet or guitar, simply play a concert A and adjust your instrument accordingly

Explanation:

It can be a bit tricky to hold a tuning fork while manipulating an instrument, which is why some musicians decide to clench the base of a ringing tuning fork in their teeth. This has the unique effect of transmitting sound through your bones, allowing your brain to "hear" the tone through your jaw. According to some urban legends, touching your teeth with a vibrating tuning fork is enough to make them explode. It's a myth, obviously, but if you have a cavity or a chipped tooth, you'll quickly find this method to be unbelievably painful.

Luckily, you can also buy tuning forks that come mounted on top of a resonator, a hollow wooden box designed to amplify a tuning fork's vibrations. In 1860, a pair of German inventors even devised a battery-powered tuning fork that musicians didn't need to ring again and again

6 0
3 years ago
Eat always flows from a place to a place.
Shkiper50 [21]

Explanation: Yes, If you digest something, it will flow from inside your neck into a tube that goes into stomach.

Hope this helps! :D

-TanqR

8 0
3 years ago
Read 2 more answers
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