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3 years ago

13

How are NGC 1427A and U different? How are they the same?

2 answers:

Lelechka [254]3 years ago

8 0

Answer: NGC 1427A has no general shape, so it is an irregular galaxy. U has a bulge in the center and arms, so it is a spiral galaxy. They are similar in that both contain plenty of dust and gas. Both also have active star-forming sites.

WINSTONCH [101]3 years ago

8 0

Answer/explanation :

NGC 1427A is an irregular galaxy in the constellation Eridanus. Its distance modulus has been estimated using the globular cluster luminosity function to be 31.01 ± 0.21 which is about 52 Mly.[2] It is the brightest dwarf irregular member of the Fornax cluster and is in the foreground of the cluster's central galaxy NGC 1399.

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The average kinetic energy of particles of matter is called

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An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force

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1) A

2) C

3) B

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5) Incomplete information(picture missing)

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9) C

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Explanation:

1) m = 15kg, a = $10ms^{-2}$ , F = ma = 15*10 = 150N

2) m = 3kg, v = $4ms^{-1}$ , r = 4m, F = $\frac{mv^{2} }{r}$

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F = $\frac{mv^{2} }{r}$ and F = ma

equating the two equations and cancelling a, we have:

$\frac{v^{2} }{r}$ = a

making v the subject of formula, we have:

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= $\sqrt{100}$

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4) r = 10m, v = $5ms^{-1}$ , a = ?

F = $\frac{mv^{2} }{r}$

F = ma

equating the above equations and making a subject of formula, we get:

a = $\frac{v^{2} }{r}$

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5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is tripled

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We can see that the force will be 9X greater than it was.

9) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is doubled

F = $(2v)^{2}$

F = 4 $v^{2}$

We can see that the force will be 4X greater than it was.

10) F = $\frac{mv^{2} }{r}$

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

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