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Reika [66]
3 years ago
14

The Milky Way is often considered to be an intermediately wound, barred spiral, which would be type ________ according to Hubble

.
Physics
1 answer:
barxatty [35]3 years ago
7 0

Answer: SBb

On 1930 the astronomer Edwin Hubble classified the galaxies based on their visual appearance into elliptical, spiral and irregular, being the first two classes the most frequent.  

So, according to this classification, the Milky Way is a barred spiral galaxy (SBb in Hubble's notation system) because it has a central bar-shaped structure of bright stars that spans from one side of the galaxy to the other. In addition, its spiral arms seem to emerge from the end of this "bar".

Scientifics considered this, after measuring the the disk and central bulge region of the galaxy, and the conclusion is the Milky Way fulfills these conditions, because is a galaxy that orbits on its same axis and with this rotation its arms are twisted in opposite directions around the mentioned axis.

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A rat runs 2m right, turns around and runs 3m left. Then goes 2m right. What is its displacement?
geniusboy [140]

Answer:

Explanation:

I think this answer would be 1m to the left.

7 0
2 years ago
1.a car sitting at rest on the road <br>balanced or unbalanced<br>explain your answer​
Paladinen [302]

Answer:

Balanced force

Explanation:

Balanced Forces, When forces are in balance, acceleration is zero. Velocity is constant and there is no net or unbalanced force. ... Although friction is acting on the person, there is no change in velocity and friction is not a net force in this case. Friction is only a net force if it changes the velocity of a mass.

7 0
3 years ago
Read 2 more answers
How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−21 n
hichkok12 [17]

Answer:

891 excess electrons must be present on each sphere

Explanation:

One Charge = q1 = q

Force = F = 4.57*10^-21 N  

Other charge = q2 =q

Distance = r = 20 cm = 0.2 m  

permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)  

Using Coulomb's law,

F=[1/4pieo]q1q2/r^2

F = [1/4pieo]q^2 / r^2

q^2 =F [4pieo]r^2

q =  r*sq rt F[4pieo]

 q=0.2* sq rt[ 4.57 x 10^-21]*[4*3.1416*8.854*10^-12]

q = 1.42614*10^ -16 C

number of electrons = n = q/e=1.42614*10^ -16 /1.6*10^-19

n =891

 891 excess electrons must be present on each sphere  

5 0
2 years ago
A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on
Juliette [100K]

Answer:

k = 5\times 10^{4}\ N/m

b = 0.707\times 10^{3}

t = 7.1\times 10^{- 5}\ s

Solution:

As per the question:

Mass of the block, m = 1000 kg

Height, h = 10 m

Equilibrium position, x = 0.2 m

Now,

The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

v^{2} = u^{2} + 2gh

where

u = initial velocity = 0

g = 10m/s^{2}

Thus

v = \sqrt{2\times 10\times 10} = 10\sqrt{2}\ m/s

Force on the mass is given by:

F = mg = 1000\times 10 = 10000 N = 10\ kN

Also, we know that the spring force is given by:

F = - kx

Thus

k = \frac{F}{x} = \frac{10000}{0.2} = 5\times 10^{4}\ N/m

Now, to find the damping constant b, we know that:

F = - bv

b = \frac{F}{v} = \frac{10000}{10\sqrt{2}} = 0.707\times 10^{3}

Now,

Time required for the platform to get settled to 1 mm or 0.001 m is given by:

t = \frac{0.001}{v} = \frac{0.001}{10\sqrt{2}} = 7.1\times 10^{- 5}\ s

4 0
3 years ago
A roller coaster car travels up and down the hills of its track. Neglecting
Volgvan

Answer:

option D is the correct option

Must always remain constant

Explanation:

According to their law of conservation of energy :it states that in a closed system,the total mechanical energy is always constant although energy may change from one form to another. e. g from potential energy to kinetic energy

7 0
3 years ago
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