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Reika [66]
3 years ago
10

Astronomers have observed a small, massive object at the center of our Milky Way galaxy. A ring of material orbits this massive

object; the ring has a diameter of about 15 light years and an orbital speed of about 200 km/s.
Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star?
Physics
1 answer:
vazorg [7]3 years ago
3 0

Answer: Yes, it can be a single, ordinary star.

Explanation: To determine a mass of a star, we use the orbital speed formula, given by: v = \sqrt{\frac{GM}{R} }, where

v is the speed;

G is a constant: G = 6.67*10^{-11}\frac{m^{3} }{kg.s^{2} }

M is mass of a massive object;

R is the distance between the object orbiting and the massive object;

The formula can be rewritten as:

M = \frac{v^{2}.R }{G}

First, we change R from light years to km:

1km=1.057*10^{-13}

R= \frac{15}{2*1.057.10^{-13} }

Calculating mass:

M = \frac{2^{2}*10^{4}*14.2*10^{13} }{6.67*10^{-11} }

M = 4.25*10^{28} kg

A solar mass is the standard unit of mass. It is approximately 2*10^{30}Kg and can be used for comparison: A single star cannot be more than 50 solar masses.

50 solar masses = 50*2*10^{30} = 10^{32} kg

Comparing the mass of the object with this parameter, we have

\frac{10^{32} }{4.25.10^{28} } = 0.235.10^{4} = 2.35.10^{3}

From this, we know that 50 solar masses is greater than the small, massive object found. So, this object can be a single, ordinary star.

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KengaRu [80]

Answer:

a) \ d_2=d_1\\b) \ d_2=4d_1

Explanation:

Assume that the distance travelled initially is d.

In order to stop the block you need some external force which is friction.

If we use the law of energy conservation:

E_i=E_f\\\frac{mv^2}{2}= E_{Friction}\\E_{Friction}=F_{Friction}*d\\F_{Friction}= \mu_kmg\\\frac{mv^2}{2}= \mu_kmgd\\ d=\frac{v^2}{2\mu_kg}

a)

Looking at the formula you can see that the mass doesn't affect the distance travelled, as lng as the initial velocity is constant (Which indicates that the force must be higher to push the block to the same speed) therefore the distance is the same.

b) If the velocity is doubled, then the distance travelled is multiplied by 4, because the distance deppends on the square of the velocity.

6 0
4 years ago
}Un alumno utiliza una balanza de dos platillos y coloca un objeto en el platillo derecho. Luego coloca las masas patrón en el i
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Answer:

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Explanation:

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6 0
3 years ago
Pls someone I need it urgently and explain Solving and explanation so I can understand Thank you
Temka [501]

Answer:

   f = 6.37 Hz,       T = 0.157 s

Explanation:

The expression you have is

       y = 5 sin (3x - 40t)

this is the equation of a traveling wave, the general form of the expression is

      y = A sin (kx - wt)

where A is the amplitude of the motion, k the wave vector and w the angular velocity

Angle velocity and frequency are related

         w = 2π f

         f = w / 2π

from the equation w = 40 rad / s

        f = 40 / 2π

        f = 6.37 Hz

frequency and period are related

       f = 1 / T

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3 years ago
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8 0
3 years ago
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An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary
Alenkinab [10]

Answer:

T =  4200N

Explanation:

When the submersible craft is at rest, the tension in the cable is 6000N.

With this information you can calculate the weight of the craft by summing the forces (the summation of the force is zero because the craft is at rest):

T-W=0\\\\W=T=6000N

When the craft is going down with a constant speed, there is a drag force of 1800N. Then, by using the second Newton law you have:

T-W+F_d=0   (1)

Fd: drag force

The summation of the forces is zero because the craft moves with constant velocity, that is, there is no acceleration.

You calculate the new tension on the cable by solving the equation (1) for T:

T=W-Fd=6000N-1800N=4200N

hence, the tension is 4200N

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