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3 years ago

An air mass that forms over Arizona and New Mexico will be?

Physics

1 answer:

vlada-n [284]3 years ago

5 0

Continental Tropical Air Mass. Your welcome matey

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How do the amplitudes of a 120 decibel sound and a 100-decibel sound compare?

NemiM [27]

The 120 decibel sound has more amplitude than the 100 decibel sound.

In Physics, the relation between amplitude and intensity is that the intensity of the wave is directly proportional to the square of its amplitude.

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2 years ago

Explain Thomsons model of an atom<br><br><br>please its aurgent fast

Aleks04 [339]

Answer:

Thomson's model showed an atom that had a positively charged medium, or space, with negatively charged electrons inside the medium. After its proposal, the model was called a "plum pudding" model because the positive medium was like a pudding, with electrons, or plums, inside.

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2 years ago

A force of 80 N is exerted on an object on a frictionless surface for a distance of 4 meters. If the object has a mass of 10 kg,

solmaris [256]

There are two ways to find energy. Energy=F*d=mv^2. We can use this relationship to find v:
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3 years ago

lcompute the increase in length of 500 m of copper wire when it's temperature changes from 12 degrees celsius to 32 degrees Cels

marshall27 [118]

Explanation:

step by step explanation

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2 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

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3 years ago