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andreev551 [17]
3 years ago
15

How does the redshift of distant galaxies best support the big bang theory? It shows that the galaxies are becoming warmer. It s

hows that the galaxies are becoming larger. It shows that the galaxies are moving farther away. It shows that the galaxies are shrinking in size.
Physics
2 answers:
Elan Coil [88]3 years ago
7 0
It shows that galaxies are moving further away
cupoosta [38]3 years ago
5 0

Answer:

It shows that the galaxies are moving farther away.

Explanation:

When light from the galaxies is shifted towards the longer wavelengths, it is called red shift. When light shifts towards the shorter wavelengths, it is called blue shift. Big bang theory states how the primordial universe started expanding from a hot and dense singularity, in the process of evolution of the universe.

The  universe is expanding always. Galaxies that are a part of the universe keep expanding and move farther away from each other. The galaxies that are very far off appear to  move much faster than the nearby galaxies.

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A force of 100 newtons Is applled to a box at an angle of 36° with the horizontal. If the mass of the box Is 25 kilograms, what
faust18 [17]
Horizontal component of force = 100cos(36)= 80.9 N
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2 years ago
See the diagram above. Which of the following is the best prediction of what would happen if you increased the distance in secti
Inessa [10]

By the definition of wavelength, the answer is the letter D, the wavelength would decrease.

We can see in the diagram a wave motion.

A wave has some characteristics:

  • Has an amplitude, the distance from 0 to the crest (highest point in the y-direction, point (3) in the figure) it would see in the figure as (2)
  • Has wavelength, the distance between the crests.
  • Has a trough, the lowest point in the y-direction.

Now, if we increase the distance of the crests, by the definition shown above, we will increase the wavelength.

Therefore, the answer is letter D, the wavelength would increase.

You can learn more about wave motion here:  

brainly.com/question/22763521

7 0
2 years ago
A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
2 years ago
During destructive interference, two waves moving through the same medium will
ddd [48]
They will amplify eachother.
6 0
2 years ago
Read 2 more answers
There is a force which resists changes in motion, called
adoni [48]

Answer:

inertia 11

Explanation:

5 0
3 years ago
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