Interstellar matter is quite evenly distributed throughout the milky way galaxy. true or false

Physics

1 answer:

melomori [17]3 years ago

4 0

Answer:

False

Explanation:

In addition to stars, our galaxy contains abundant diffuse matter that is distributed throughout its volume and constitutes what we call the interstellar medium. This medium plays a fundamental role in the life cycle of the stars, since it is where the matter from which they are born resides, and it is the place to which it returns when the stars expel their outer layers at death.

The interstellar medium is a complex environment. Its matter is not distributed uniformly, but consists of different phases with temperatures ranging from a few degrees Kelvin (near absolute zero) in the areas of star formation to the millions of degrees Kelvin observed in supernova remnants. The densities of interstellar matter also vary orders of magnitude according to the phase, but they are always so low that they rival those that can be achieved in the best vacuum chambers of terrestrial laboratories. Depending on the density and temperature conditions, interstellar matter is in a molecular, atomic, or ionized state, although the state is not permanent, since matter circulates between the different phases in a continuous cycle of evolution on a galactic scale.

Due to the very different characteristics of its multiple phases, the interstellar medium has to be studied using various observational techniques and different types of telescopes. The coldest components of the interstellar medium do not emit visible light, and require the observation of telescopes sensitive to the weak emission of radio waves that this material produces. Using different radio telescopes, such as the 40-meter diameter of the Yebes Observatory, which the Institute of Radio Astronomy Millimeter, to which the IGN belongs, has in Grenoble and Granada, or the recently opened Atacama Large Millimeter / submillimeter Array in the Atacama desert in Chile, astronomers from the National Astronomical Observatory contribute to characterize the physical and chemical properties of the molecular clouds where stars are born and of the circumestellar shells produced by the stars in the last stages of their lives . The study of these regions is helping to complete our knowledge of the most unknown phases of the complex life cycle of stars.

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Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 2

Sidana [21]

Answer:

The work done on the wagon is 37 joules.

Explanation:

Given that,

The force applied by Charlie to the right, F = 37.2 N

The force applied by Sara to the left, F' = 22.4 N

We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :

$F_n=F-F'$

$F_n=37.2-22.4$

$F_n=14.8\ N$

Work done on the wagon is given by the product of net force and displacement. It is given by :

$W=F_n\times d$

$W=14.8\ N\times 2.5\ m$

W = 37 Joules

So, the work done on the wagon is 37 joules. Hence, this is the required solution.

7 0

2 years ago

Read 2 more answers

Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air

valentina_108 [34]

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: $v_f=v_0+a*t$ , where $v_f$ is the final velocity, $v_0$ is the initial velocity, $a$ the acceleration, and $t$ is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: $t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s} }{-201\frac{m}{s^2} }=1.40s$ , where $v_f=0\frac{m}{s}$ because the sled is totally stopped, $v_0=282\frac{m}{s}$ is the velocity of the sled before braking and, $a=-201\frac{m}{s^2}$ is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration: $x_f-x_0=v_0*t+\frac{1}{2}*a*t^2$ , where $x_f-x_0$ is the distance traveled, $v_0$ is the initial velocity, $t$ the time of the process and, $a$ is the acceleration of the process.