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DerKrebs [107]
3 years ago
10

How was Edwin Hubble able to use his discovery of Cepheids in Andromeda to prove that the "spiral nebulae" were actually galaxie

s external to the Milky Way?
Physics
1 answer:
Ad libitum [116K]3 years ago
7 0

Answer:  The correct answer is :  From the period-luminosity relation for Cepheids, he was able to determine the distance to Andromeda and show that it was far outside the Milky Way Galaxy.

Explanation:  Hubble's law says that the recession velocity of a galaxy is directly proportional to its distance from us. Hubble measured the distance to the Andromeda galaxy by applying the period-luminosity relationship to Cepheid.

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Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates <0, 4.00 cm> and <0, -4.00 cm
Dmitry [639]

Answer:

Explanation:

Potential energy of the system of charges

=  9 x 10⁹ x [  q₁q₂ / r₁₂ +  q₂q₃ / r₂₃ +  q₁q₃ / r₁₃ ]

here  r₁₂ ,  r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.

r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C

Potential energy  =  9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08  +  -200x10⁻¹⁸ / .04 +  200 x 10¹⁸ / .04 ]

= 9 x 10⁹  x  - 400 x 10⁻¹⁸ / .08

= 45 x 10⁻⁶ J .

b)

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

= 9 x 10⁹ x 10 x 10⁻⁹ / .03

= 3000 V .

potential energy of fourth particle = charge x potential

= 3000 x 40 x 10⁻⁹ = 12 x 10⁻⁵ J .

kinetic energy at infinity = 12 x 10⁻⁵ J

1/2 m v² = 12 x 10⁻⁵ J

.5 x 2 x 10⁻¹³ x v² = 12 x 10⁻⁵

v² = 12 x 10⁸

v = 3.46 x 10⁴ m/s

= 9 x 10⁹

5 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> Which number marks a crest?
lyudmila [28]

number 2 because the curve demstrates the crest GOOD LUCK i hope i got you the correct answer if not sorry

5 0
3 years ago
Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?
MrMuchimi
 T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r 

where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance. 
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect, 

Point 1: 
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r

Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m 
The distance between the two points then is equal to 7.07 m.


</span>
8 0
2 years ago
A girl is sitting in a sled sliding horizontally along some snow (there is friction present). The mass of the girl is 29.8 kg an
jonny [76]

Answer:

28.81 m

Explanation:

Ff = -123

m * a  = -123

(29.8+10.3) * a = -123

a = -123/40.1 = -3.07

We know,

v^2 = u^2 + 2as

0^2 = 13.3^2 + 2*(-3.07)*s

s = 176.89/6.14 = 28.81

[ If there's a problem with the solution, pleaase let me know ]

6 0
3 years ago
Which of the following statements about iron filings placed upon glass resting on top of a bar magnet is false? A. The direction
IceJOKER [234]
<span>C. The filings will be clustered more densely where the field is weakest.</span>
3 0
3 years ago
Read 2 more answers
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