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; 
Age of the universe; 
We know that, the reciprocal of the Hubble's constant (
) gives an estimate of the age of the universe (
). It is expressed as:

Now,
Hubble's constant; 
We know that;

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;

Now, we input this Hubble's constant value into our equation;

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Learn more: brainly.com/question/14019680
Enclosed is some guidance algebra.I find this q a little confusing. It quotes "RC" which usually makes me think of electrical circuits and time constants based on converting calculating RC value and equating that to t for one time constant then 2RC for two time constants etc. The theory being that after 5 time constants - 5RC - a circuit is stable. BUT, this q then goes on to mention HALF LIFE. The curves for both half life and time constant are both exponential, as in the number e to the power of something, but the algebra is slightly different. I hope my algebra is ok.
The first thing you should know for this case is that work is defined as the product of force by the distance traveled in the direction of force.
We have then:
W = Fd
The distance varies, so we must integrate:
from 0 to 20:
W = ∫F (x) dx
W = ∫32xdx
W = 32∫xdx
W = 32 (x ^ 2/2) = (16) (20 ^ 2) = 6400 ft * lbs
answer:
6400 ft * lbs is work done pulling the rope up 20 ft
The force that keeps the puck moving is 0.25 N while the velocity of the puck is 3.7 m/s.
<h3>What is the centripetal force?</h3>
We know that the centripetal force is the force that acts on a body that is moving along a circular path. In this case, we are told that the puck is moving along a circular path hence it is acted upon by the centripetal force that acts on it.
The centripetal force in this case would be supplied by the weight of the object that is moving in the circular path. Thus we can write in our equation that;
Centripetal force = Weight of object = mg
m = mass of the object
g = acceleration due to gravity
Then;
W = 0.026 Kg * 9.8 m/s^2
W = 0.25 N
To obtain the velocity of the object;
FT = mv^2/r
v = √ FT r/m
v = √0.25 * 1.4/0.026
v = 3.7 m/s
Learn more about centripetal force:brainly.com/question/11324711
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