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3 years ago

3. What can you infer about the position of the galaxies 100 million years before this telescope photo was taken

1 answer:

8 0

The galaxies are in constant motion so it can be inferred that their position will not be the same as a hundred million years ago.

A galaxy is a term to refer to the set of stars, gas clouds, planets, cosmic dust, dark matter, and energy gravitationally united in a more or less defined structure in the universe.

Galaxies are in constant motion because the elements that make them up are not static at any time; The galaxies rotate around the center of the galaxy. If they were still, the gravitational attraction would cause them to immediately fall towards the center of the galaxy: it is the same that would happen to the Earth and the other planets if they stopped rotating around the Sun, they would fall towards the Sun.

According to the above, it can be inferred that the galaxies after a while have moved from the place where they were last seen.

Note: This question is incomplete because there is some missing information. However, I can answer based on my general prior knowledge.

Learn more in: brainly.com/question/15451060

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A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t

Marianna [84]

Mass of the block = 1.4 kg

Weight of the block = mg = 1.4 × 9.8 = 13.72 N

Normal force from the surface (N) = 13.72 N

Acceleration = 1.25 m/s^2

Let the coefficient of kinetic friction be μ

Friction force = μN

F(net) = ma

μmg = ma

μg = a

μ = $\frac{a}{g}$

μ = $\frac{1.25}{9.8}$

μ = 0.1275

Hence, the coefficient of kinetic friction is: μ = 0.1275

6 0

3 years ago

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

pantera1 [17]

Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$

The sum of the vertical components of F1 and F2 must equal the total force Ft

$\displaystyle -2F\ cos21^o+F\ sin\alpha =460$

Solving for $\alpha$ in the first equation

$\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o$

$\displaystyle cos\alpha =0.717=>\alpha =44.214^o$

$\displaystyle F(2cos21^o+sin\alpha)=460$

$\displaystyle F=\frac{460}{2cos21^o+sin\alpha}$

$\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}$

$\displaystyle F=179.37\ N$

The magnitude of F1 is

$|F1|=2*F=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

4 0

3 years ago

How can a organization encourage total person development among their employees.

Sedbober [7]

Answer:

Paying for employees seminars and workshops related to their careers

Explanation:

To motivate personal development among employees, several things can be done. Among them, giving employees chance to present their own solutions to problems, exposing the employees to several global challenges and how to handle them, paying for employees seminars and workshops related to their own careers for professional development among other things.

8 0

3 years ago

Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109

r-ruslan [8.4K]

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

$E=\Delta mc^2\\\Delta m=\frac{E}{c^2}$ (1)

That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

$E=Pt=2.3*10^{9}\frac{J}{s}*1 year*\frac{365.25 day}{1 year}*\frac{24h}{1 day}*\frac{3600s}{1h}=7.25*10^{16}J$

Hence, by replacing in the equation (1) you have (c=3*10^{8}m/s)

$\Delta m=\frac{7.25*10^{16}J}{3*10^{8}\frac{m}{s}}=2.41*10^{8}kg$

HOPE THIS HELPS!!

3 0

3 years ago

Read 2 more answers

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road