Question

problem set #10 begin date: 11/22/2025 12:01:00 am due date: 12/4/2025 11:59:00 pm end date: 12/7/2025 11:59:00 pm problem 13: (6% 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 760 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 = 1.672×10⁴ km/s correct! - 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, the number is a rough estimate for the age of the universe. t = 4.2×10¹⁷ s correct! - part (c) convert your answer from part (b) to years. t = yr hints: 1% deduction per hint. hints remaining: 1 10 submission(s) remaining submit hint feedback: 1% deduction per feedback.

Explanation:

Step1: Recall conversion factor

We know that there are approximately \(3.154\times10^{7}\) seconds in one year. So the conversion factor from seconds to years is \(\frac{1\ \text{yr}}{3.154\times 10^{7}\ \text{s}}\).

Step2: Multiply the time in seconds by the conversion factor

We are given \(t = 4.200\times10^{17}\ \text{s}\). To convert to years, we calculate \(t_{\text{yr}}=4.200\times 10^{17}\ \text{s}\times\frac{1\ \text{yr}}{3.154\times 10^{7}\ \text{s}}\)

First, divide the coefficients: \(\frac{4.200}{3.154}\approx1.332\)

Then, divide the powers of 10: \(10^{17 - 7}=10^{10}\)

Multiply these results together: \(1.332\times 10^{10}\approx1.332\times10^{10}\) (we can also calculate it more precisely: \(4.200\times10^{17}\div(3.154\times 10^{7})=\frac{4.200}{3.154}\times10^{17 - 7}\approx1.332\times 10^{10}\))

Answer:

\(1.332\times 10^{10}\) (or approximately \(1.33\times 10^{10}\) or more precise value depending on the calculation of \(\frac{4.200}{3.154}\))