Question

final exam
begin date: 12/5/2025 12:01:00 am due date: 12/13/2025 11:59:00 pm end date: 12/13/2025 11:59:00 pm
problem 23: (9% of assignment value)
one way to derive the rough age of the universe is to use the value of the hubble constant and a reliable distance to a far - away galaxy (with the assumption that the value of the hubble constant has not changed since the big bang).

• part (a) ✓

consider a galaxy at a distance of 778 million light - years receding from us at velocity v. if the hubble constant is 22 km/s per million light - years, what is its velocity in km/s?
v = 4.3×10¹⁷ km/s
v = 4.300×10¹⁷ km/s
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• part (b) ✓

how long ago (in seconds) was that galaxy right next door to our own galaxy if it has always been receding at its present rate? since the universe began when all galaxies were very close together, this number is a rough estimate for the age of the universe.
t = 1.36×10¹⁰
t = 1.360×10¹⁰ s
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• part (c)

convert your answer from part (b) to years.
t =
yr
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Explanation:

Step1: Recall conversion factors

We know that 1 year has approximately \(3.154\times10^{7}\) seconds (or we can use the given conversion factor approach: 1 year = 365 days, 1 day = 24 hours, 1 hour = 3600 seconds, so \(1\ \text{year}=365\times24\times3600 = 31536000\approx3.15\times10^{7}\) seconds). We are given \(t = 1.360\times10^{10}\) seconds.

Step2: Perform the conversion

To convert seconds to years, we divide the number of seconds by the number of seconds in a year. Let \(t_{\text{years}}=\frac{t_{\text{seconds}}}{\text{seconds per year}}\). Using the approximation of \(3.15\times10^{7}\) seconds per year:

\(t_{\text{years}}=\frac{1.360\times10^{10}}{3.15\times10^{7}}\)

First, divide the coefficients: \(\frac{1.360}{3.15}\approx0.4317\)

Then, divide the powers of 10: \(\frac{10^{10}}{10^{7}} = 10^{3}=1000\)

Multiply these results: \(0.4317\times1000 = 431.7\) (using more precise calculation: \(\frac{1.360\times10^{10}}{3.1536\times10^{7}}=\frac{1.360}{3.1536}\times10^{3}\approx0.4312\times10^{3}=431.2\))

If we use the exact calculation from the given (maybe the problem expects using \(1\ \text{year}=3.154\times10^{7}\) or the calculator's input), but following the steps:

\(t = 1.360\times10^{10}\ \text{s}\)

Seconds in a year: \(365\times24\times3600 = 31536000 = 3.1536\times10^{7}\ \text{s/year}\)

\(t_{\text{years}}=\frac{1.360\times10^{10}}{3.1536\times10^{7}}=\frac{1.360}{3.1536}\times10^{3}\approx0.4312\times10^{3}=431.2\) (or more accurately, let's compute \(\frac{1.360\times10^{10}}{3.1536\times10^{7}}\))

\(1.360\times10^{10}\div(3.1536\times10^{7})=(1.360\div3.1536)\times10^{3}\approx0.4312\times1000 = 431.2\) (rounding as needed, maybe the answer is approximately 431 or 432, but let's check with the given value in part (b) which was \(1.36\times10^{10}\) s)

Wait, the problem says "Convert your answer from part (b) to years". From part (b), the time was \(t = 1.360\times10^{10}\) s.

So using the conversion:

\(1\ \text{year}=3.1536\times10^{7}\ \text{s}\)

So \(t_{\text{years}}=\frac{1.360\times10^{10}\ \text{s}}{3.1536\times10^{7}\ \text{s/year}}\)

Calculate the division:

\(1.360\times10^{10}\div3.1536\times10^{7}=\frac{1.360}{3.1536}\times10^{3}\)

\(1.360\div3.1536\approx0.4312\)

\(0.4312\times10^{3}=431.2\)

So approximately 431 years? Wait, no, wait, \(10^{10}\div10^{7}=10^{3}=1000\), and \(1.36\div3.15\approx0.43\), so \(0.43\times1000 = 430\) or 431.

But let's do it more accurately:

\(3.1536\times10^{7}\ \text{s/year}\)

\(1.360\times10^{10}\ \text{s}\div(3.1536\times10^{7}\ \text{s/year})=(1.360\div3.1536)\times10^{3}\)

\(1.360\div3.1536 = 13600\div31536\approx0.4312\)

\(0.4312\times1000 = 431.2\)

So the answer is approximately 431 years (or more precisely, around 431).

Answer:

\(\boxed{431}\) (or a more precise value like 431.2 depending on the required precision)