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3 years ago

Which galaxy type is smooth and uniform in appearance, where no new stars are forming?

Physics

1 answer:

Oksi-84 [34.3K]3 years ago

6 0

Answer: Elliptical galaxy

Explanation:

There are mainly three types of galaxies: Spiral, Elliptical and Irregular. Elliptical galaxies do not have any arms unlike spiral galaxies. The elliptical galaxies are smoother and uniform in appearance. These contain relatively old stars with any site of new star formation missing. In fact, scientists consider elliptical galaxies as those galaxies which are nearing the end of their evolution.

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Any fracture or system of fractures along which Earth moves is known as a

azamat

The answer is B.) Stress

7 0

2 years ago

Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The

tia_tia [17]

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

Charge will be distributed in the ratio of their capacity

charge on sphere1 = q x 1 / ( 1 + 4.5 )

= q / 5.5

fraction = 1 / 5.5

c ) charge on sphere 2

= q x 4.5 / 5.5

fraction = 4.5 / 5.5

d ) surface charge density of sphere 1

= q /( 5.5 x A ) where A is surface area

surface charge density of sphere 2

= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area

= q /( 5.5 x 4.5 A )

q_1/q_2 = 4.5

6 0

2 years ago

2. An athlete of average size is hanging from the end of a 20 m long rope, which has a mass of 4 kg and is attached to a hook in

a_sh-v [17]

Answer:

t = 0.319 s

Explanation:

With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by

v = √T/λ

Linear density is

λ = m / L

λ = 4/20

λ = 0.2 kg / m

The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg

T = W = mg

T = 80 9.8

T = 784 N

The pulse rate is

v = √(784 / 0.2)

v = 62.6 m / s

The time it takes to reach the hook can be searched with kinematics

v = x / t

t = x / v

t = 20 / 62.6

t = 0.319 s

7 0

3 years ago

Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What

vesna_86 [32]

Answer:

$1.02\cdot 10^{-8} N$ , repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have same sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the two electrons

$r=1.5\cdot 10^{-10} m$ is their separation

Substituting into the formula, we find the electric force between them:

$F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N$

And the force is repulsive, since the two electrons have same sign charge.

4 0

2 years ago

A heat engine does 200 j of work per cycle while exhausting 600 j of heat to the cold reservoir. what is the engine's thermal ef

AveGali [126]

The thermal efficiency of an engine is
$\eta= \frac{W}{Q}$
where
W is the work done by the engine
Q is the heat absorbed by the engine to do the work

In this problem, the work done by the engine is W=200 J, while the heat exhausted is Q=600 J, so the efficiency of the machine is
$\eta= \frac{W}{Q}= \frac{200 J}{600 J}=0.33 = 33 \%$

8 0

3 years ago