All categories

3 years ago

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:

You might be interested in

If a toy has a mass of 1 kg, it has an Earth weight of

marshall27 [118]

Answer:

The answer is 9.8 N

Explanation:

As we know that the weight of an object is the amount of gravitational force acting on the object in an upward direction if the weight is acting is the downward direction.

The formula of weight:

W = Mass x Gravitational force

W = m x g

Given data:

Mass =1 kg

g = 9.8 ms-2

W = 1kg x 9.8 ms-2 = 9.8 kgms-2 ( 1 kgms-2 = N)

SO,

W = 9.8 N

The toy has an earth weight of 9.8 N.

4 0

3 years ago

RACTIC PTUDIES

Amanda [17]

Answer:

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

Explanation:

The distance of the book before the lamp is moved, $d_{b} = 30 cm$

The distance of the book after the lamp is moved, $d_{a} = 90 cm$

Illumination can be given by the formula, $E = \frac{P}{4 \pi d^{2} }$

Illumination before the lamp is moved, $E_{b} = \frac{P}{4 \pi d_{b} ^{2} }$

Illumination after the lamp is moved, $E_{a} = \frac{P}{4 \pi d_{a} ^{2} }$

$\frac{E_{a}}{E_{b}} } = \frac{\frac{P}{4 \pi d_{a} ^{2} } }{\frac{P}{4 \pi d_{b} ^{2} } }$

$\frac{E_{a} }{E_{b} } = \frac{d_{b} ^{2} }{d_{a} ^{2}} \\\frac{E_{a} }{E_{b} } = \frac{30^{2} }{90 ^{2}}\\\frac{E_{a} }{E_{b} } =\frac{900 }{8100}\\\frac{E_{a} }{E_{b} } =\frac{1 }{9}$

$E_{b} = 9E_{a}$

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

6 0

2 years ago

You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler

scZoUnD [109]

Answer:

q=1.4*10^{-9}C

Explanation:

Given data:

charge on ruler = -14μC

Mass of tissue is 5 g

To Know the minimum charge, equate electrostatic force to weight

we have F = W

so $\frac{KQq}{r^2} =mg$

putting all value in equation,

$=\frac{9*10^9*(14*10^{-6})*q}{0.06^2} = 5* 10^{-3}*9.8$

solving for q

$q =\frac{5* 10^{-3}*9.8 *0.06^2}{9*10^9*(14*10^{-6})}$

or q=1.4*10^{-9}C

5 0

3 years ago

A group of scientists have obtained some experimental results.

topjm [15]

<span>In order best to find out whether the obtained experimental results are worth mor etime and resources the group of scientists should present their results (could be done also in poster session) to other scientists in order to hear their opinion and get a feedback. The shoudl also ask another researchers to redo the experiments and to compare the results. </span>

4 0

3 years ago

Read 2 more answers

A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the

Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

$qvB = m\frac{v^2}{r}$