<span>To know if there were other factors that affected the volume of a gas, Genaris and her classmates should: </span>"formulate a new hypothesis with the same dependent variable but a different independent variable as the original hypothesis." In this case, the dependent variable is the volume of the gas and the new independent variable is a factor they think will affect the volume of the gas.
P=4800kgm/s
As
p=mΔv
where p is momentum, m is mass and v is velocity
Given values is
m =1200kg
Δv= 17m/s-13m/s=4m/s
Now
p=mΔv
p=(1200kg)*(4m/s)
p=4800kgm/s
Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Given the data in the question;
Hubble's constant; ![H_0 = 51km/s/Mly](https://tex.z-dn.net/?f=H_0%20%3D%2051km%2Fs%2FMly)
Age of the universe; ![t = \ ?](https://tex.z-dn.net/?f=t%20%3D%20%5C%20%3F)
We know that, the reciprocal of the Hubble's constant (
) gives an estimate of the age of the universe (
). It is expressed as:
![Age\ of\ Universe; t = \frac{1}{H_0}](https://tex.z-dn.net/?f=Age%5C%20of%5C%20Universe%3B%20t%20%3D%20%5Cfrac%7B1%7D%7BH_0%7D)
Now,
Hubble's constant; ![H_0 = 51km/s/Mly](https://tex.z-dn.net/?f=H_0%20%3D%2051km%2Fs%2FMly)
We know that;
![1\ light\ years = 9.46*10^{15}m](https://tex.z-dn.net/?f=1%5C%20light%5C%20years%20%3D%209.46%2A10%5E%7B15%7Dm)
so
![1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m](https://tex.z-dn.net/?f=1%5C%20Million%5C%20light%5C%20years%20%3D%20%5B9.46%20%2A%2010%5E%7B15%7Dm%5D%20%2A%2010%5E6%20%3D%209.46%20%2A%2010%5E%7B21%7Dm)
Therefore;
![H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\](https://tex.z-dn.net/?f=H_0%20%3D%2051%5Cfrac%7Bkm%7D%7B%5Cfrac%7Bs%7D%7BMly%7D%20%7D%20%3D%2051000%5Cfrac%7Bm%7D%7Bs%5C%20%2A%5C%20Mly%7D%20%20%5C%5C%5C%5CH_0%20%3D%2051000%5Cfrac%7Bm%7D%7Bs%5C%20%2A%5C%20%289.46%2A10%5E%7B21%7Dm%29%7D%20%5C%5C%5C%5CH_0%20%3D%20%205.39%20%2A10%5E%7B-18%7Ds%5E%7B-1%7D%5C%5C)
Now, we input this Hubble's constant value into our equation;
![Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years](https://tex.z-dn.net/?f=Age%5C%20of%5C%20Universe%3B%20t%20%3D%20%5Cfrac%7B1%7D%7BH_0%7D%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7B1%7D%7B%205.39%20%2A10%5E%7B-18%7Ds%5E%7B-1%7D%7D%20%5C%5C%5C%5Ct%20%3D%201.855%20%2A%2010%5E%7B17%7Ds%5C%5C%5C%5CWe%5C%20convert%5C%20to%5C%20years%5C%5C%5C%5Ct%20%3D%20%20%5Cfrac%7B%201.855%20%2A%2010%5E%7B17%7D%7D%7B60%2A60%2A24%2A365%7Dyrs%20%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7B%201.855%20%2A%2010%5E%7B17%7D%7D%7B31536000%7Dyrs%5C%5C%5C%5Ct%20%3D%205.88%20%2A10%5E9%20years)
Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Learn more: brainly.com/question/14019680