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Maslowich
3 years ago
15

Whats an astral projection

Physics
2 answers:
Dominik [7]3 years ago
5 0

Answer:Esotericism

Explanation:

it’s something that’s in intentional out of body experience

uranmaximum [27]3 years ago
5 0

Answer:

Astral projection is a term used in esotericism to describe an intentional out-of-body experience that assumes the existence of a soul or consciousness called an "astral body" that is separate from the physical body and capable of travelling outside it throughout the universe.

Explanation:

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You observe a distant galaxy. You find that a spectral line of hydrogen that is shifted from its normal location in the visible
dimulka [17.4K]

Answer:

The galaxy is moving away from us

Explanation:

Galaxy refers to a system of millions or billions of stars accompanied by gas and dust such that they are held together as a result of gravitational attraction.

When we observe a distant galaxy, we find that a spectral line of hydrogen that is shifted from its normal location in the visible part of the spectrum into the infrared part of the spectrum.

It means that the galaxy is moving away from us.

6 0
3 years ago
What force is necessary to accelerate 32kg object by 2 m/s
Oxana [17]

Answer:

i8nnndnsjdjcicicixidocin where dosoldmxbeus sosojsnnd

Explanation:

nzjwiemddsisjxnxn,jisiskxnxjxjxiisnz MN nsjs. snxjsisisis s. xjsisisk

6 0
2 years ago
Pluto was first observed in 1930, and its largest moon, Charon, was discovered in 1978. A few years after Charon’s discovery, as
mylen [45]

Answer:

The average densities of both matches the expected density for objects made from water ice.

Explanation:

Charon's density is 1.2 to 1.3 g / cm3, while Pluto's density is 1.8 to 2.1 g / cm3. This was discovered in many researches and measurements of these two celestial bodies, with the objective of understanding them and promoting efficient scientific knowledge.

With the measurements of the average densities between pluto and Charon it was possible to conclude several statements about them. Firstly, it is possible to see that the two formed independently and at different times, in addition to indicating the existence of few rocks in charon, which is consistent with the average density of objects made mostly of water ice.

8 0
3 years ago
"1.0 kg mass is attached to the end of a spring. The mass has an amplitude of 0.10 m and vibrates 2.0 times per second. Find its
prisoha [69]

Answer:

1.3m/s

Explanation:

Data given,

Mass,m=1.0kg,

Amplitude,A=0.10m,

Frequency,f=2.0Hz.

From the equation of a simple harmonic motion, the displacement of the object at a given time is define as

x=Acos\alpha \\

we can express the velocity by the derivative of the displacement,

Hence

V=-Awsin\alpha \\

at equilibrium, the velocity becomes

V=wA\\w=2\pi f

Hence if we substitute values we arrive at

V=2\pi fA\\V=2\pi *2*0.1\\V=1.3m/s

4 0
3 years ago
Read 2 more answers
When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequenc
Nesterboy [21]

Answer:

w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

Explanation:

We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is

          w² = mg d / I

In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow

         d = L / 2

The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated

        I = ¼ m r2 + ⅓ m L2

        I = m (¼ r2 + ⅓ L2)

now let's use the concept of density to calculate the mass of the system

        ρ = m / V

        m = ρ V

the volume of a cylinder is

         V = π r² L

          m =  ρ π r² L

let's substitute

        w² = m g (L / 2) / m (¼ r² + ⅓ L²)

        w² = g L / (½ r² + 2/3 L²)

        L >> r

         w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

4 0
4 years ago
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