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34kurt
3 years ago
10

What is the correct answer?

Physics
1 answer:
inysia [295]3 years ago
3 0
The answer to your question is metaphase
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The cosmic microwave background essentially looks the same in all directions. This is an example of.
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In general, all views of the cosmic microwave background are identical. Isotropy is demonstrated by this.

<h3>What exactly is isotropy, for instance?</h3>

The Greek words isos (equal) and tropos, from which the term "isotropy" is derived, mean "uniform in all directions" (way). The material properties of anisotropic materials, such as graphite, differ depending on the direction, in contrast to isotropic materials like glass, which show the same properties in all directions.

There is no "centre" to an isotropic universe, which is another characteristic. The North and South Poles are produced by the rotation of the Earth, giving them a distinctive orientation, but the Universe is visible from every angle. When we think about the Big Bang, which is the origin of the Universe, this is a crucial point.

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A ball is thrown vertically upwards with a velocity of 30m/s. Determine the maximum height reached
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The maximum height reached is 45.92 m

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9. What do people playing pool use to determine their shots?
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An imaginary line perpendicular to a reflecting surface is called _________.
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3 years ago
Read 2 more answers
A uniform solid disk rolls without slipping down an incline making an angle θ with the horizontal. What is its acceleration? (En
Maru [420]

Answer:

aCM = (2/3)*g*Sin θ

Explanation:

Consider a uniform solid disk having mass M,  radius R and rotational inertia I  about its center of mass, rolling without  slipping down an inclined plane.

In order to get the linear acceleration of the object’s center of mass, aCM ,

down the incline,  we analyze this as follows:

The force of gravity (W = Mg) acting straight down  is resolved into components parallel and  perpendicular to the incline.

Since the object rolls without  slipping there is a force of  friction (Ff) acting on the object,  at it’s point of contact with the  incline, in the direction up  the incline.

Newton’s 2nd Law gives then for acceleration down the incline

∑Fx' = m*aCM   ⇒    m*g*Sin θ - Ff = m*aCM

The force of friction also causes a torque around the center of mass

having lever arm R so we can also write

τ = R*Ff = I*α

Solving for the friction,    Ff = I*α / R

This is used in the expression  derived from the 2nd Law:

m*g*Sin θ - Ff = m*g*Sin θ - (I*α / R) = m*aCM

The objects angular acceleration is related to the linear acceleration  of the edge that contacts the incline by

a = R*α

Since the object rolls without  slipping this has the same  magnitude as aCM so we have  that

α = aCM / R

Using this in

m*g*Sin θ - (I*α / R) = m*g*Sin θ - (I*(aCM / R) / R) = m*aCM

⇒  aCM = (m*g*Sin θ*R²) / (I + m*R²)

if I = (1/2)*m*R²   (for a uniform solid disk)

we get

aCM = (2/3)*g*Sin θ

6 0
3 years ago
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