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nasty-shy [4]
1 year ago
10

what is the designation given to a galaxy with a small nuclear bulge and loosely wound arms starting from a a bar through the nu

clear bulge
Physics
1 answer:
Naya [18.7K]1 year ago
5 0

In SBc galaxies, both the arms and the bar are highly resolved into star clouds and stellar associations. The arms are open in form and can start either at the ends of the bar or tangent to a ring.

<h3>Is the Milky Way a SBc galaxy?</h3>

Originally thought to be a hubble type Sb or Sc (Sbc), astronomers now believe that the Milky Way is has a central bar and is therefore a loosely wound, barred spiral galaxy – Hubble type SBbc.

<h3>What is the difference between an SBc galaxy and an Sc galaxy?</h3>

Sa (and SBa) galaxy types have tightly wound arms and large central bulges.

Sc's (and SBc's) have loosely wound arms and a small central bulge.

Learn more about SBc galaxy here:

<h3>brainly.com/question/2515850</h3><h3 /><h3>#SPJ4</h3>

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In  Medicine, "etiology" means the origin or cause of a disease.

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3 years ago
Calcula qué tan lejos está Saturno del Sol, sabiendo que la luz solar necesita una hora y veinte segundos para llegar a Saturno.
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Answer:

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Explanation:

6 0
3 years ago
Determine the current in the 7-ohm resistor for the circuit shown in the figure. Assume that the batteries are ideal and that al
seraphim [82]

Answer:

I₁ = 1.6 A (through 7 Ohm Resistor)

I₂ = 1.3 A (through 8 Ohm Resistor)

I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)

Explanation:

Here we consider two loops doe applying Kirchhoff's Voltage Law (KVL). The 1st loop is the left side one with a voltage source of 12 V and the 2nd Loop is the right side one with a voltage source of 9 V. We name the sources and resistor's as follows:

R₁ = 7 Ω

R₂ = 4 Ω

R₃ = 8 Ω

V₁ = 12 V

V₂ = 9 V

Now, we apply KVL to 1st Loop:

V₁ = I₁R₁ + (I₁ - I₂)R₂

12 = 7I₁ + (I₁ - I₂)(4)

12 = 7I₁ + 4I₁ - 4I₂

I₁ = (12 + 4 I₂)/11   ------------ equation (1)

Now, we apply KVL to 2nd Loop:

V₂ = (I₂ - I₁)R₂ + I₂R₃

9 = (I₂ - I₁)(4) + 8I₂

9 = 4I₂ - 4I₁ + 8I₂

9 = 12I₂ - 4I₁   -------------- equation (2)

using equation (1)

9 = 12I₂ - 4[(12 + 4 I₂)/11]

99 = 132 I₂ -  48 - 16 I₂

147 = 116 I₂

I₂ = 147/116

I₂ = 1.3 A

use this value in equation 2:

9 = 12(1.3 A) - 4I₁

4I₁ = 15.6 - 9

I₁ = 6.6 A/4

I₁ = 1.6 A

Hence, the currents through all resistors are:

<u>I₁ = 1.6 A (through 7 Ohm Resistor)</u>

<u>I₂ = 1.3 A (through 8 Ohm Resistor)</u>

<u>I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)</u>

4 0
3 years ago
All the bulb in the series circuit are on or off together?
emmainna [20.7K]
Yes they are the power goes from one to another and if one goes off another will go off too
5 0
3 years ago
The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through alumi
Alex777 [14]

Answer:

(A) The correct answer is option (B) three halves that of the old unit.

(B) The answer is three fourth that of old unit

Explanation:

from the relation;

(A) Fromthe expression;

K = Qd/AΔT

Anew = 3/2 A(old)

(B)

K¹ = 2K(old), so we get

A(old) = A(old)/2

Combining with part A, we have;

Anew = 3/2 *A(old)/2

          = 3/4A(old)

The answer is three fourth that of old unit

8 0
3 years ago
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