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3 years ago

In which era did the universe’s clouds start to condense and the universe became transparent for the first time?

Physics

1 answer:

alukav5142 [94]3 years ago

8 0

Answer:

The Universe became transparent to the light left over from the Big Bang when it was roughly 380,000 years old

Explanation:

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If an atom has 13 protons and is currently electrically neutral, what must happen to give the same atom a positive charge of +2e

Alex777 [14]

The atom must lose 2 electrons

Explanation:

An atom consists of three types of particles:

- Protons, in the nucleus, with positive electric charge $+e$

- Neutrons, in the nucleus, with no electric charge

- Electrons, orbiting around the nucleus, with negative electric charge $-e$

As a result, the net electric charge of an atom is given by the number of protons minus the number of electrons:

Q = #p - #e

For a neutral atom, the number of protons and electrons is the same, so the net charge is zero.

In order for an atom to have a positive charge of +2, it means that there must be 2 protons more than the number of electrons. Since atoms exchange electrons (and not protons), this means that the atom must have "lost" 2 electrons.

In this problem, we have an atom with 13 protons: this means that initially it also has 13 electrons. However, later the atom lost 2 electrons, and as a result, the final charge is +2e.

Learn more about atoms:

brainly.com/question/2757829

#LearnwithBrainly

5 0

4 years ago

An indicator of average kinetic energy is _____. pressure volume diffusion temperature

slega [8]

An indicator of average kinetic energy is temperature. Temperature is directly proportional to Kinetic energy of the molecules of an element.

3 0

3 years ago

Read 2 more answers

A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carouse

Sidana [21]

Answer:

m = 35.98 Kg ≈ 36 Kg

Explanation:

I₀ = 125 kg·m²

R₁ = 1.50 m

ωi = 0.600 rad/s

R₂ = 0.905 m

ωf = 0.800 rad/s

m = ?

We can apply The law of conservation of angular momentum as follows:

Linitial = Lfinal

⇒ Ii*ωi = If*ωf <em>(I)</em>

where

Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m

If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m

Now, we using the equation <em>(I) </em>we have

(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800

⇒ m = 35.98 Kg ≈ 36 Kg

5 0

3 years ago

Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3

VARVARA [1.3K]

Answer:

Part a)

$f = 4.76 \times 10^{14} Hz$

Part b)

$d = 3.48 \times 10^{-4} m$

Part c)

$\theta = 0.311 degree$

Explanation:

Part a)

As we know that the speed of light is given as

$c = 3 \times 10^8 m/s$

$\lambda = 630 nm$

now the frequency of the light is given as

$f = \frac{c}{\lambda}$

so we have

$f = \frac{3 \times 10^8}{630 \times 10^{-9}}$

$f = 4.76 \times 10^{14} Hz$

Part b)

Position of Nth maximum intensity on the screen is given as

$y_n = \frac{n\lambda L}{d}$

so here we know for 3rd order maximum intensity

$y_3 = 0.76 cm$

n = 3

L = 1.4 m

$0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}$

$d = 3.48 \times 10^{-4} m$

Part c)

angle of third order maximum is given as

$d sin\theta = 3 \lambda$

$3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})$

$\theta = 0.311 degree$

8 0

3 years ago

If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol

xenn [34]

Answer:

Average density of Sun is 1.3927 $\frac{g}{cm}$ .

Given:

Radius of Sun = 7.001 × $10^{5}$ km = 7.001 × $10^{10}$ cm

Mass of Sun = 2 × $10^{30}$ kg = 2 × $10^{33}$ g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Solution:

Density of Sun is given by,

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Volume of Sun = $\frac{4}{3} \pi r^{3}$

Volume of Sun = $\frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}$

Volume of Sun = 1.436 × $10^{33}$ $cm^{3}$

Density of Sun = $\frac{ 2\times 10^{33} }{1.436 \times 10^{33} }$

Density of Sun = 1.3927 $\frac{g}{cm}$

Thus, Average density of Sun is 1.3927 $\frac{g}{cm}$ .

4 0

3 years ago