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3 years ago

What can quantum physics tell us about the nature of our reality?

1 answer:

7 0

As do pixels create such a defined picture of virtual reality it would be similar to say that such small particles (with understanding and exploring them) would show us how things interact and that basic interaction might show us the result of the creating of something from which we can discover by observing the final result of that partical or even as if looking at a cake- to soon know what it takes to make a cake maybe by it's flavor or how it's not yet known "ingredients" interact with the known ingredients and that interaction shows that ingredients to the cake and the ingredients might as well be the future of that cake

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Any help is appreciated

sdas [7]

image distance,di=10 cm

object distance,do=20cm

magnification, m=di/do

=10/20

=0.5

since the image is virtual, magnification is negative.

therefore m=-0.5

8 0

2 years ago

ფიზიკაში მჭირდება დახმარება ნაგლმა მასწავლებელმა მთელი თვე გააცდინა და როცა დაბრუნდა ეგრევე საკონტროლო ისეთი საკითხებით რომელიც

MakcuM [25]

Answer:

The photo is blocked cant see anything

Explanation:

3 0

2 years ago

If the earth can travel 1800 km in 1 minute what is it’s speed in km/s

Lady bird [3.3K]

Answer:

30km/s

Explanation:

1800/60=30

4 0

3 years ago

Which statement best describes the main idea of the poem

SSSSS [86.1K]

Explanation:

option D ) is correct the speaker explain a difficult decision he had to make

hi everyone comment

7 0

2 years ago

When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c

e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = $e^{kt}$

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = $e^{-0.01t}$

V = V₀ $e^{-0.01t}$

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀ $e^{-0.01t}$

0.1V₀ = V₀ $e^{-0.01t}$

0.1V₀/V₀ = $e^{-0.01t}$

0.1 = $e^{-0.01t}$

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0

3 years ago