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irinina [24]
3 years ago
8

What is a lenticular (S0) galaxy? A galaxy with a disk and central bulge like a spiral galaxy, but with no spiral arms A galaxy

with a smooth brightness profile, lacking the central bulge and disk of a spiral galaxy A spiral galaxy with fuzzy and poorly formed spiral arms A galaxy with a lot of gas and dust and no particular structure
Physics
1 answer:
RSB [31]3 years ago
6 0

Answer:

A galaxy with a disk and central bulge like a spiral galaxy, but with no spiral arms

Explanation:

A Lenticular galaxy is a kind of galaxy intermediate between elliptical galaxy and a spiral galaxy in the Morphological classification system of galaxies. They have a central bulge or disc just like a Spiral galaxy but lacks the arms of spiral galaxy. If looked edge on they appear to be spiral and if looked face on they appear to be elliptical.

The absence of spiral arms can be attributed to the absence of star formation. They mainly consists of ageing stars.

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What are subtomic particles<br>​
slavikrds [6]

Answer:

In the physical sciences, subatomic particles are smaller than atoms. They can be composite particles, such as the neutron and proton; or elementary particles, which according to the standard model are not made of other particles. Particle physics and nuclear physics study these particles and how they interact.

Explanation:

3 0
2 years ago
Read 2 more answers
If a dog is mass iS 14.3 kg what is it’s weight on earth
BigorU [14]

Answer:

Weight of the dog on surface of earth is 140.14 Newton.

Given:

mass of the dog = 14.3 kg

To find:

Weight of the dog = ?

Formula used:

Weight of the dog is given by,

W = mg

Where, W = weight of the dog

m = mass of the dog

g = acceleration due to gravity

Solution:

Weight of the dog is given by,

W = mg

Where, W = weight of the dog

m = mass of the dog = 14.3 kg

g = acceleration due to gravity

W = 14.3 × 9.8

W = 140.14 Newton

Weight of the dog on surface of earth is 140.14 Newton.


8 0
3 years ago
An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m
Tomtit [17]

Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

Find: a and t

(47.5)² = (34.3)² + 2a(40100)

a = 0.0135 m/s²

40100 = ½ (47.5 + 34.3)t

t = 980 s

7 0
3 years ago
A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.
Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we  can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

y=(u sin \theta)t-\frac{1}{2}gt^2 (1)

where:

u sin \theta is the initial vertical velocity of the shell, with u=156 ft/s and \theta=49.0^{\circ}

g=32 ft/s^2 is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

x=(ucos \theta)t

where u cos \theta is the initial horizontal velocity of the shell.

We can re-write this last equation as

t=\frac{x}{u cos \theta} (1b)

And substituting into (1),

y=xtan\theta -\frac{1}{2}gt^2 (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of \alpha=-41.0^{\circ} from the horizontal, so we can write the position (x,y) of the hill as

y=x tan \alpha (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

xtan\alpha = xtan \theta - \frac{1}{2}gt^2

Substituting (1b) into this equation,

xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0

Which has 2 solutions:

x = 0 (origin)

and

tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft

So, the distance d down the hill at which the shell strikes the hill is

d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

t=\frac{x}{u cos \theta}

where:

x = 1322 ft is the final position of the shell when it strikes the hill

u=156 ft/s is the initial velocity of the shell

\theta=49.0^{\circ} is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s

Instead, the vertical component of the velocity is given by

v_y=usin \theta -gt

And substituting at t = 12.9 s (time at which the shell strikes the hill),

v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s

Therefore, the  final speed of the shell is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0
3 years ago
A proton in a particle accelerator is traveling at a speed of 0.99c.(a) If you use the approximate nonrelativistic equation for
____ [38]

Answer:

a)  p = 4.96 10⁻¹⁹ kg m / s , b)  p = 35 .18 10⁻¹⁹  kg m / s ,

c)  p_correst / p_approximate = 7.09

Explanation:

a) The moment is defined in classical mechanics as

                 p = m v

Let's calculate its value

               p = 1.67 10⁻²⁷ 0.99 3. 10⁸

               p = 4.96 10⁻¹⁹ kg m / s

b) in special relativity the moment is defined as

               p = m v / √(1 –v² / c²)

Let's calculate

                p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)

                p = 4.96 10⁻¹⁹ / 0.141

                p = 35 .18 10⁻¹⁹  kg m / s

c) the relationship between the two values ​​is

            p_correst / p_approximate = 35.18 / 4.96

            p_correst / p_approximate = 7.09

4 0
3 years ago
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