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3 years ago

How did the Planck mission help scientists learn about the age of the universe

Physics

1 answer:

forsale [732]3 years ago

3 0

Answer:

By using observations from the Atacama Cosmology Telescope (ACT) in Chile, the new findings match the measurements of the Planck satellite data of the same ancient light.

Explanation:

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When rockets split do they come back to earth?

attashe74 [19]

I'm not 100% sure, but I believe what you mean is when they eject the old propulsion motors. Yes, they land in the ocean and the US Navy retrieves them for later use.

4 0

3 years ago

Part 1

xz_007 [3.2K]

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ .

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ .

<h3>Determination of the coordinates of the center of gravity</h3>

Let suppose that each square has an <em>uniform</em> density, the coordinates of the center of gravity of each square with respect to the origin are, respectively:

$\vec r_{1} = (0.5\cdot a, 0.5\cdot a)$

$\vec r_{2} = (1.5\cdot a, 0.5\cdot a)$

$\vec r_{3} = (0.5\cdot a, 1.5\cdot a)$

$\vec r_{4} = (1.5\cdot a, 1.5\cdot a)$

The center of gravity of the <em>entire</em> system is found by applying definition of <em>weighted</em> averages:

$\vec r_{cg} = \frac{W_{1}\cdot \vec r_{1}+W_{2}\cdot \vec r_{2}+W_{3}\cdot \vec r_{3}+W_{4}\cdot \vec r_{4}}{W_{1}+W_{2}+W_{3}+W_{4}}$

Where $W_{1}$ , $W_{2}$ , $W_{3}$ and $W_{4}$ are weights of the each square, in newtons.

Now we proceed the coordinates of the center of gravity of the entire system:

$\vec r_{cg} = \frac{(50\,N)\cdot (0.5\cdot a, 0.5\cdot a) + (30\,N)\cdot (1.5\cdot a, 0.5\cdot a)+(30\,N)\cdot (0.5\cdot a, 1.5\cdot a)+(70\,N)\cdot (1.5\cdot a, 1.5\cdot a)}{50\,N+30\,N+30\,N+70\,N}$

$\vec r_{cg} = \frac{5}{18}\cdot (0.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (1.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (0.5\cdot a, 1.5\cdot a) + \frac{7}{18}\cdot (1.5\cdot a, 1.5\cdot a)$

$\vec r_{cg} = \left(\frac{19}{18}\cdot a, \frac{19}{18}\cdot a \right)$ $\blacksquare$

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ . $\blacksquare$

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ . $\blacksquare$

To learn more on center of gravity, we kindly invite to check this verified question: brainly.com/question/20662119

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2 years ago

EASY. PLEASE HELP. David is on a school bus stopped at a pick up point where students are getting on the bus. Once everyone is s

goldenfox [79]

Answer:

Explanation:

because when things go forward quickly things get pushed back

8 0

3 years ago

Who was the first man to walk on the moon?

Andrej [43]

The first person is Neil Armstrong

8 0

4 years ago

A potential energy function for a system in which a two-dimensional force acts is of the form U = 3x6y − 5x. Find the force that

liq [111]

Answer:

Force, $F=((-18x^5y+5)i+(-3x^6)j)\ N$

Explanation:

Given that,

A potential energy function for a system in which a two-dimensional force acts is of the form of :

$U=3x^6y-5x$

We need to find the force that acts at the point (x, y). The force in 2 dimensional with components is given by :

$F=(\dfrac{-dU}{dx},\dfrac{-dU}{dy})\\\\F=(\dfrac{-d(3x^6y-5x)}{dx},\dfrac{d(3x^6y-5x)}{dy})\\\\F=(-(18x^5y-5)),(-(3x^6))\\\\F=((-18x^5y+5)i+(-3x^6)j)\ N$

So, the force acting at the point (x,y) is $((-18x^5y+5)i+(-3x^6)j)\ N$ . Hence, this is the required solution.

7 0

3 years ago