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Roman55 [17]
3 years ago
6

What kind of stars make up the galactic nucleus?

Physics
2 answers:
Solnce55 [7]3 years ago
5 0
The galactic nucleus or the active galactic nucleus (AGN) is the region of the galaxy that is considered compact. It's luminosity is higher than normal which is not generated by stars. Therefore, the kind of stars that composes the galactic nucleus are the old, metal-rich stars.
Hatshy [7]3 years ago
4 0
I think the correct answer would be old and metal poor stars are found in the galactic nucleus. This nucleus us a region in the center of a galaxy which contains a higher luminosity than other parts. It produces very high amounts of energy. Hope this helps.
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Ryan swings a pail of water in a vertical circle 1.0 m in radius at a constant speed. If the water is NOT to spill on him:
nexus9112 [7]

Part 1

If water does not spill at the top point of the circular motion then for the minimum speed condition we can say normal force will be zero at the top position

F_g = ma

mg = m\frac{v^2}{R}

g = \frac{v^2}{R}

v = \sqrt{Rg}

given that

R = 1 m

g = 9.8 m/s^2

now from above equation we have

v = \sqrt{1(9.8)} = 3.13 m/s

Part b)

for minimum value of angular speed we will have

\omega = \frac{v}{R}

\omega = \frac{3.13}{1}

\omega = 3.13 rad/s

3 0
3 years ago
A hot-air balloon plus cargo has a mass of 308 kg and a volume of 2910 m3 on a day when the outside air density is 1.22 kg/m3. T
lesya [120]

9514 1404 393

Answer:

  1.114 kg/m³

Explanation:

The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...

  2910d +308 = 2910×1.22

Solving for d, we find ...

  2910d = 2919(1.22) -308

  d = 1.22 -308/2910

  d ≈ 1.114 . . . kg/m³

The density of the hot air is about 1.114 kg/m³.

6 0
3 years ago
A car of mass 500kg travelling at 60m/s has it speed reduced to 40m/s by a constant breaking force over a distance of 200m. find
uranmaximum [27]

Answer:

Ek1 = 900000 [J]

Ek1 = 400000 [J]

Explanation:

In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

E_{k1}=\frac{1}{2}*m*v1^{2}

where:

m = mass = 500 [kg]

v1 = 60 [m/s]

So we have:

Ek1 = 0.5*500*(60^2)

Ek1 = 900000 [J]

and:

Ek2 = 0.5*500*(40^2)

Ek2 = 400000 [J]

6 0
3 years ago
Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that hav
Vikki [24]

Answer:

The force constant is  k =1.316 *10^{7} \  N/m

The energy stored in the spring is  E =  1.68 *10^{7} \ J

Explanation:

From the question we are told that

   The mass of the object is  M  = 4.8*10^{5} \ kg

    The period is T  = 1.2 \ s

The period of the spring oscillation is  mathematically represented as

         T  =2 \pi \sqrt{ \frac{M}{k}}

where  k is the force constant

   So making k the subject

       k = \frac{4 \pi ^2 M }{T^2}

substituting values

       k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}

      k =1.316 *10^{7} \  N/m

The energy stored in the spring is mathematically represented  as

       E =  \frac{1}{2} k x^2

Where x is the spring displacement which is given as

        x =  1.6 \ m

substituting values

      E =  \frac{1}{2} (1.316 *10^{7}) (1.6)^2

       E =  1.68 *10^{7} \ J

   

7 0
3 years ago
Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts
Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground  = 3.13 m/s

4 0
3 years ago
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