What two things make up the 96% of the universe

1 answer:

5 0

I know that dark matter, things astronomers are unable to see even with the highest of current technology, makes up most of the universe.

Most of the universe is made up of stars and galaxies, but if you wanna go even further into that, it's filled with gas and dust.

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The only major country that does not use the metric system

horrorfan [7]

It’s for sure United States (B)

3 0

3 years ago

4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with

serious [3.7K]

Answer:

Option A is the correct answer.

Explanation:

Here momentum is conserved.

That is $\left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}$

Substituting values

$3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s$

Speed of block A after collision = 10 m/s

Option A is the correct answer.

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3 years ago

Oil flows upward in the wick of a lantern because of the liquid property called

vlabodo [156]

Oil flows upward in the wick of a lantern because of the liquid property called <span>Capillarity c:</span>

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3 years ago

1. Astronomers' ever-expanding knowlage of the universe has coinsided with the development of _____, a collection of indirect di

DerKrebs [107]

The best and most correct answer among the choices provided by your question is the first choice or letter A.

<span> Astronomers' ever-expanding knowledge of the universe has coincided with the development of the cosmic distance scale, a collection of indirect distance measurement.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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3 years ago

Read 2 more answers

10%) Problem 7: Water flows through a water hose at a rate of Q1 = 620 cm3/s, the diameter of the hose is d1 = 1.99 cm. A nozzle

Semenov [28]

Answer:

a) $A_1 = \frac{\pi d_1^2}{4}$

Explanation:

a) the cross-sectional area of the hose would be the square of radius times pi. And since the sectional radius is half of its diameter d. We can express the cross-sectional area A1 in term of diameter d1

$A_1 = \pi r_1^2 = \pi (d_1/2)^2 = \frac{\pi d_1^2}{4}$

6 0

3 years ago