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3 years ago

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According to the Big Bang theory, after the "Bang", the universe remained dark until about ___ later, when neutral atoms began f

orm.
*three minutes
*one billion years
*300,000 years
*three billion years

2 answers:

fiasKO [112]3 years ago

5 0

300,000 Years is the right answer Your Welcome

barxatty [35]3 years ago

3 0

300,000 years this i know because i had to do a report on it last year and it says in the description/ study article of the big bang theory.

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A certain circuit is composed of two series resistors. The total resistance is 10 Ohms. One of the resistors is 4 Ohms. The othe

Mariulka [41]

<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>

A certain circuit is composed of two series resistors
The total resistance is 10 ohms
One of the resistor is 4 ohms

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

We have to find the value of other resistor?

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>

We know that,

In series combination,

When a number of resistances are connected in series, the equivalent I.e resultant resistance is equal to the sum of the individual resistances and is greater than any individual resistance

<u>That </u><u>is</u><u>, </u>

Rn in series = R1 + R2 + R3.....So on

<u>Therefore</u><u>, </u>

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

We have,

R1 + R2 = 10 Ω

4 + R2 = 10Ω

R2 = 10 - 4

R2 = 6Ω

Hence, The value of R2 resistor in series is 6Ω

4 0

2 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago

‘The Universe is expanding.’ Justify this statement with the Big bang model of universe and Hubble’s Law.

Taya2010 [7]

Explanation:

The expansion of the universe is the increase in distance between any two given ... Metric expansion is a key feature of Big Bang cosmology, is modeled mathematically with the ...

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2 years ago

Read the scenario below and answer the question that follows. Jane is a psychologist. She submits a study on sleep to a prestigi

Setler79 [48]

A:the other options are examples of unethical behavior

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3 years ago

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When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner

Paul [167]

Answer:

1353.38 Watt

Explanation:

T₁ = Initial temperature of the house = 35°C

T₂ = Final temperature of the house = 20°C

Δt = Time taken to cool the house = 38 min = 38×60 = 2280 s

m = mass of air in the house = 800 kg

Cv = Specific heat at constant volume = 0.72 kJ/kgK

Cp = Specific heat at constant pressure = 1.0 kJ/kgK

Heat removed

q = mCvΔT

⇒q = 800×720×(35-20)

⇒q = 8640000 J

Average rate of hear removal

$Q=\frac{q}{\Delta t}\\\Rightarrow Q=\frac{8640000}{2280}\\\Rightarrow Q=3789.47\ W$

$COP=\frac{Q}{W}\\\Rightarrow W=\frac{Q}{COP}\\\Rightarrow W=\frac{3789.47}{2.8}\\\Rightarrow W=1353.38\ W$

∴ Power drawn by the air conditioner is 1353.38 Watt

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3 years ago

Read 2 more answers