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elixir [45]
3 years ago
8

When astronomers observe the spectra of distant galaxies,they notice that the hydrogen emission lines are shifted noticeably tow

ard the red end of the visual spectrum,a phenomenon called red shift. Red is lowest frequency of visible light. What does red shift indicate about the movement of the distant galaxies?
Physics
1 answer:
34kurt3 years ago
5 0
The amount of redshift increases in relation with the distance, meaning, the larger the redshift, the more distant the galaxy.  The Hubble diagram which was created by Edwin Hubble in 1929 shows that the more redshifted a galaxy is the further away it is. The galaxies are moving away from Earth because the fabric of space itself is expanding.  <span>Subtle changes in the color of starlight let </span>astronomers<span> find planets,  </span>measure<span> the speeds of </span>galaxies<span>, and track the expansion of the universe. </span>
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A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and
vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

U=mgh

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

h=4-4cos(30\°)=0.53m

Then, the potential energy is:

U=(55kg)(9.8m/s^2)(0.53m)=285.67J

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

U'=(m_1+m_2)gh'=285.67\ J

From the previous equation you can get h':

h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

8 0
3 years ago
A block with a mass of 8.7 kg is dropped from rest from a height of 8.7 m, and remains at rest after hitting the ground. 1)If we
Harlamova29_29 [7]

To solve this problem we will apply the concepts related to gravitational potential energy.

This can be defined as the product between mass, gravity and body height.

Mathematically it can be expressed as

\Delta P = mgh

\Delta P = (8.7)(9.8)(3)

\Delta P = 255.78J

Therefore the change in the internal energy of the system is 255.78

7 0
3 years ago
Help when u can! I feel like it's C, but I just want to make sure before I submit the test.
Marat540 [252]

(D)

Explanation:

This kind of bond where the electrons are transferred is called ionic bond. This bond occurs between a metallic and non-metallic elements. The other kind of bond (covalent) involves the sharing of electrons and this happens between non-metallic elements.

4 0
2 years ago
A dart gun shoots a dart with an angle of 45' above horizontal During the upward part of the trajectory the gravitational accele
ludmilkaskok [199]

Answer:

4. Downward and its value is constant

Explanation:

As this is a case of projectile motion, we use the reference frame where upward direction to be positive for y, and in the same way to be negative in the downward direction. On another hand, we have that gravity is always acting this means that gravitational acceleration g is directed downward constantly over the dart not only during the upward but also during the downward part of the trajectory. And it is ruled by the following equations.

For the x-axis

v_{x}=v_{0}cos(45\°)=constant

x=(v_{0}cos(45\°))t

For the y-axis

v_{y}=v_{0}sin(45\°)-gt

y=v_{0}sin(45\°)t-\frac{1}{2} gt^{2}

Where v_{0}, is the initial velocity.

8 0
3 years ago
THE NUMERICAL RATIO OF DISPLACEMENT TO DISTANCE IS:​
Tamiku [17]

There's no general rule.  

Displacement is the length of a straight line from start to finish, and distance is how far you actually traveled from start to finish.

The only thing we really know is that distance can never be shorter than displacement.  So I guess the answer is:

<em>The numerical ratio of displacement to distance is always 1 or less</em>.

(But it has to be written ALL IN CAPITALS.)

7 0
3 years ago
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