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Andrej [43]
2 years ago
15

Do we always see the same side of the moon from earth?.

Physics
1 answer:
Sergio [31]2 years ago
6 0
The Moon orbits Earth once every 27.3 days and spins on its axis once every 27.3 days. This means that although the Moon is rotating, it always keeps one face toward us. Known as “synchronous rotation,” this is why we only ever see the Moon's nearside from Earth.
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When kids construct a whole world around something they believe, it is called ____.
Oksana_A [137]
The answer is emagination emagination
8 0
3 years ago
Read 2 more answers
Two beams of coherent light are shining on the same piece of white paper. with respect to the crests and troughs of such waves,
defon

Answer:

"where crests and troughs have their maxima at the same time"

Crests and troughs are 180 deg out of phase and when they have their maxima at the same time and place, their net contribution will be zero"

7 0
2 years ago
A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse t
WARRIOR [948]

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

          θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

         θ = 1.22 550 10⁻⁹ / 0.002

         θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

       tan θ = y / L

       y = L tan θ

       y = 2,389 10⁵ tan 3,355 10⁻⁴

       y = 8.02 10¹ mi

       y = 80.2 mille

This is the smallest size of an object seen directly by the eye

5 0
3 years ago
Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1
Artemon [7]

Answer:

\frac{F_1}{F_2} =\frac{1}{2}

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

F_1=2k\times d

case 2:

F_2=2k\times 2d

therefore, \frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}

\frac{F_1}{F_2} =\frac{1}{2}

4 0
3 years ago
A cart with mass m vibrating at the end of a spring has an extra block added to it when its displacement is x=+A. What should th
Pavel [41]

Answer:

The block's mass should be 3m

Explanation:

Given:

Cart with mass m

From the conservation of energy before mass is added,

  \frac{1}{2} mv^{2} = \frac{1}{2} kA^{2}

Where A = amplitude of spring mass system, k = spring constant

  A = v\sqrt{\frac{m}{k} }

Now new mass M is added to the system,

   \frac{1}{2} (m +M ) v^{2}  = \frac{1}{2}  k A^{2}

  A = v \sqrt{\frac{m +M }{k} }

Here, given in question frequency is reduced to half so we can write,

   f' = \frac{f}{2}

Where f = frequency of system before mass is added, f' = frequency of system after mass is added.

        \omega ' = \frac{\omega}{2}

\sqrt{\frac{k}{m +M} }  = \frac{\sqrt{\frac{k}{m} } }{2}

   \frac{k}{m +M } = \frac{k}{4m}

   M = 3m

Therefore, the block's mass should be 3m

8 0
3 years ago
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