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lys-0071 [83]
3 years ago
6

Most of the bright stars in our galaxy are located in the galactic

Physics
1 answer:
GenaCL600 [577]3 years ago
3 0

False. Most of the bright stars in our galaxy are located in the disk.

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This is one popular brand of exercise machine for a crossword puzzle
Rudiy27

Answer:

Aerobics I think.

Explanation:

7 0
2 years ago
Question: A 0.50 m wave has a frequency of 686 Hz. What is the speed of the wave?
Dafna1 [17]
C. 340 
Frequency is the number of wavelengths per second and since the length is 0.5 you multiply 0.5*686 and get 343.
the question not allowing for one position thus the answer is c

7 0
3 years ago
Which of the following most easily transfers charge?
raketka [301]
A. Conductors. Because in conductors, gap between valence band and conduction band is negligible. Due to this electrons from valence band moves spontaneously to conduction band and on small heating the conductors.

3 0
3 years ago
Read 2 more answers
Earth is 149.6 million meters from the Sun and takes 365 days to make one complete revolution around the Sun. Mars is 227.9 mill
luda_lava [24]

Answer:

\frac{a_{r,earth}}{a_{r,mars}} = 2.325

Explanation:

The distance of Earth from the Sun is 149.6\times 10^{9}\,m and of Mars from the Sun is 227.9\times 10^{9}\,m. Let assume that both planets have circular orbits. The centripetal accelaration can be found by using the following expression:

a_{r} = \frac{v^{2}}{R}

Since planet has translation at constant speed, this formula is applied to compute corresponding speeds:

v=\frac{2\pi\cdot r}{\Delta t}

Earth:

v_{earth} = \frac{2\pi\cdot (149.6\times 10^{9}\,m)}{(365\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}

v_{earth}=29806.079\,\frac{m}{s}

Mars:

v_{mars} = \frac{2\pi\cdot (227.9\times 10^{9}\,m)}{(687\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}

v_{mars}=24124.244\,\frac{m}{s}

Now, centripetal accelarations can be found:

Earth:

a_{r,earth} = \frac{(29806.079\,\frac{m}{s} )^{2}}{149.6\times 10^{9}\,m}

a_{r,earth} = 5.939\times 10^{-3}\,\frac{m}{s^{2}}

Mars:

a_{r,mars} = \frac{(24124.244\,\frac{m}{s} )^{2}}{227.9\times 10^{9}\,m}

a_{r,mars} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}

The ratio of Earth's centripetal acceleration to Mars's centripetal acceleration is:

\frac{a_{r,earth}}{a_{r,mars}} = \frac{5.939}{2.554}

\frac{a_{r,earth}}{a_{r,mars}} = 2.325

7 0
3 years ago
Can someone please answer these in the next 20 minutes. ill give you a brainiest. thanks! :)
oee [108]

Answer:

1. 2380m

2. 13.6m

3. 132 Hz

6 0
3 years ago
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