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ASHA 777 [7]
3 years ago
7

If Hubble's constant is 73 km/s/Mpc then the age of the universe is 13.4 billion years, assuming a uniform expansion. Suppose it

were discovered that Hubble's constant is actually larger than 73 km/s/Mpc. What effect would this have on the calculated age of the universe?
Physics
1 answer:
ki77a [65]3 years ago
3 0

Answer:

It would decrease the calculated age of the universe

Explanation:

Since the age of the universe is the reciprocal of Hubble's constant, it's therefore if Hubble's constant is increased the age decreases but if the Hubble's constant is decreased, the age of universe increases. Therefore, the age of universe and Hubble's constant are inversely proportional. Conclusively, any attempt to increase Hubble's constant would imply the calculated age of the universe decreases.

You might be interested in
What is the process that changes smaller atoms into larger ones in a star
Alisiya [41]

Answer:

Nuclear fusion

Explanation:

In fusion, many nuclei (the centers of atoms) combine together to make a larger one (which is a different element). ... A star's mass determines what other type of nucleosynthesis occurs in its core (or during explosive changes in its life cycle).

5 0
3 years ago
What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 43
kiruha [24]

Answer:

The magnification is m  = 0.3674

Explanation:

From the question we are told that

  The  power of the lens is  P = -4.00 D(dioptre)

Generally  1 dioptre = 1 \ meter

  The object distance is u =  -43 \ cm the negative sign is because the distance is measured in the opposite direction of incident light (i.e away )

 Generally the focal length is mathematically represented as

          f = \frac{1}{P}  

   =>f = \frac{1}{4.00 }  

  =>  f = 0.25 \ m

converting to  cm  

 =>   f = 0.25 \ m = 0.25 * 100 = 25 \ cm

Generally from lens equation  we have that  

     \frac{1}{f} +\frac{1}{v} -\frac{1}{u}

=>  \frac{1}{25} +\frac{1}{v} -\frac{1}{-43}

=>   v =  -15.8 \ cm

Generally the magnification is mathematically represented as

      m  = \frac{v}{u}

=>    m  = \frac{- 15.8}{-43}

=>    m  = 0.3674

6 0
3 years ago
Which of the following is not a way to make exercising on a cold day safer?
IRISSAK [1]

Answer:

D.

Explanation:

I think its D. because Tea during a cold workout isnt safe at all. Drinking tea while working out adds weight and you just simply dont need tea when working out.. Hope this helps man :)

4 0
3 years ago
Read 2 more answers
Marie is puzzled by her findings she has done several meticulous calculations and has gotten the numbers .37 rad, .89 rad and 1.
Irina-Kira [14]

Answer:

I think the answer is a

Explanation:

for it to be accurate has be to exactly 0.9 rad

it is not precise because the answer she is getting is different everytime and not even close. For instance,

It would have been precise if she had gotten 0.37 rad in every attempt. or 0.89 every attempt...

6 0
3 years ago
IV) Fill in the blanks:
Vlada [557]

1.) equal volume of different substances have "different" masses.

2.)The more closely packed arrangement the particles of a substance have, "increases" its density.

3.)the SI unit of power is "Watts".

4.)an iron nail sinks in water but floats on " mercury ".

5.)balloons used for advertisements are filled with " helium" gas.

6.)"Conduction" is the primary mode of heat transfer in liquid and gases.

I hope this helps you...

5 0
3 years ago
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