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Lilit [14]
3 years ago
8

If cats always land on their feet, how do they get hurt when they fall?

Physics
1 answer:
Ulleksa [173]3 years ago
6 0
The impact front the fall, depending on the distance, has an effect on the impact of their bones crushing with the pressure from the fall.
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The conductors that carry the current to electrical devices and ? equipment are the heart of all electrical systems. There are a
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Answer:

Utilization, effects

Explanation:

The conductors that carry the current to electrical devices and utilization equipment are the heart of all electrical systems. There are associated effects whenever current flows through a conductor.

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Hii can someone please doo this! 50 pointss.
andrey2020 [161]

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The periodic table illustrate some of the elements from Hydrogen to Calcium

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Velocity can best be described by which of the following statements?
pogonyaev

Answer: Velocity can best be described as, the speed in a given direction.

Explanation: To find the answer, we need to know more about the Velocity of a body.

<h3>What is Velocity of a body?</h3>
  • Velocity is the rate of change of displacement.
  • It's a vector quantity and is measured in m/s.
  • It can be positive, negative or zero.
  • A body is said to be in uniform motion, then its velocity remains constant.
  • Change in velocity can be a change in speed.
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8 0
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Select the statement that is not true about the Hubble Telescope:
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Answer:

The option is B is not true for Hubble telescope.

4 0
3 years ago
A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is
harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

Where,

d_g = Depth of glass

n_w = Refraction index of water

n_g = Refraction index of glass

n_{air} = Refraction index of air

d_w = Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})

d'w = 4.041cm

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0
3 years ago
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