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Feliz [49]
3 years ago
12

The idea that the universe began from a single point and expanded to its current size explains a large number of observations, i

ncluding those in the table below. In addition, many predictions based on the idea have led to additional observations that support it. This idea of the origin of the universe is best classified as a
Physics
2 answers:
murzikaleks [220]3 years ago
6 0
The description of the question provided above points out to the famous Big Bang Theory. In addition, this theory is among the most accepted by cosmologists because it fits like a glove to the phenomenon the universe is experiencing right now: it is expanding and distances between celestial bodies are getting farther and farther.
Zepler [3.9K]3 years ago
6 0

The origin of universe is best classified by the Big bang theory. This theory proposes that the universe is very old and its origin took place almost 20 billion years ago. It states that a single huge explosion i.e. Big bang took place in the space which was so powerful that it is unimaginable in physical terms. Due to this explosion, the universe expanded in volume and the temperature of the space came down. Slowly hydrogen and helium gases formed, which condensed due to the gravitational forces present in the surroundings. This all resulted in the formation of billions of galaxies.

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Select the decimal that is equivalent to 29/37​
morpeh [17]

Answer:

0.78

Explanation:

6 0
3 years ago
What is frequency? <br><br> (P.S. please help)
kvasek [131]

Explanation:

the rate at which something occurs or is repeated over a particular period of time or in a given sample

4 0
3 years ago
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A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straig
inna [77]

Answer:

\mu_s=\frac{1}{3}\tan \theta

Explanation:

Let the minimum coefficient of static friction be \mu_s.

Given:

Mass of the cylinder = M

Radius of the cylinder = R

Length of the cylinder = L

Angle of inclination = \theta

Initial velocity of the cylinder (Released from rest) = 0

Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.

We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,

\tau =I\alpha

Now, angular acceleration is given as:

\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}

Also, moment of inertia for a cylinder is given as:

I=\frac{MR^2}{2}

Therefore, the torque acting on the cylinder can be rewritten as:

\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1

Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are mg\sin \theta\ and\ f. The net force acting along the incline is given as:

F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2

Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.

So, N=Mg\cos \theta

Plugging in N=Mg\cos \theta in equation (2), we get

F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3

Now, as per Newton's second law,

F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4

Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.

\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times  R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)

Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:

\mu_s=\frac{1}{3}\tan \theta

3 0
3 years ago
Read 2 more answers
Star a has a parallax angle of 0.02 arcseconds, and star b has a parallax of angle of 0.1 arcseconds. which star is more distant
Studentka2010 [4]
<span>Star a is more distant and is approximately 5 times as far away as star b Parallax is the change in angle that one must do in order to observe the same object from different locations. The further away an object is, the smaller the parallax is. As the angles approach zero, the trig functions tend to be fairly linear. And 0.1 arc seconds and 0.02 arc seconds are close enough to zero for this to hold true. Since the parallax for star a is smaller than the parallax for star b, it is the more distant star. And since 0.1 divided by 0.02 = 5, it is approximately 5 times further away than star b.</span>
3 0
3 years ago
How to delete question​
Marrrta [24]
Oh your from the other question you made I just saw it LOL.
But heres the answer click the 3 dots on the question you made or you can ask a Moderator or Administrator to remove your question with a reason.
6 0
3 years ago
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