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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.

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A material through which electrons can move easily is a what
kirill [66]

Answer:

Electrical conductors

3 0
1 year ago
Can someone please help me ASAP!
Artist 52 [7]
I’m not 100% sure but i think it’s A because if you divide the speed by the time you get 2 and also all the other answer choices don’t make any sense!
3 0
3 years ago
What are the products of a fusion reaction? Check all that apply.
Free_Kalibri [48]

Energy, heavier atoms atoms, neutron and a proton are the products of a fusion reaction.

<h3>What are the products of a fusion reaction?</h3>

Nuclear fusion is a type of reaction in which two or more atomic nuclei fuse or combine to form heavy nuclei and subatomic particles such as neutrons or protons and release of energy.

So we can conclude that Energy, heavier atoms, neutron and a proton are the products of a fusion reaction.

Learn more about reaction here: brainly.com/question/11231920

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8 0
2 years ago
Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.
olga2289 [7]

Answer:

NO_{1.499}

Explanation:

Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:

n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }

n_{N} = 2.632\,kmol

n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }

n_{O} = 3.946\,kmol

Ratio of kilomoles oxygen to kilomole nitrogen is:

n^{*} = \frac{3.946\,kmol}{2.632\,kmol}

n^{*}= 1.499

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

NO_{1.499}

8 0
4 years ago
Read 2 more answers
B. Jerome plays middle linebacker for South's varsity football team. In a game against
weqwewe [10]

Answer:

Option D

670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

Final momentum is also mv but v being westward direction, we take it negative

Final momentum=82*-2.5= -205 Kg.m/s

Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s

Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s

4 0
3 years ago
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