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ryzh [129]
2 years ago
7

Diamonds are a very dense material. Predict what would happen to the light ray if you projected it from air through a diamond

Physics
2 answers:
Anni [7]2 years ago
5 0

Answer:

a cat could eat it

Explanation:

Fluffy ate my diamond

Kipish [7]2 years ago
4 0

Answer:

It [the diamond] would act like a prism, and make a rainbow, or, the light would break up and disappear

Explanation:

that's what I would think at least

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State the success of j j Thompson model theory​
vovikov84 [41]

Answer:

Thomson's atomic model was successful in explaining the overall neutrality of the atom. However, its propositions were not consistent with the results of later experiments. In 1906, J. J. Thomson was awarded the Nobel Prize in physics for his theories and experiments on electricity conduction by gases.

Summary. J.J. Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons. Thomson proposed the plum pudding model of the atom, which had negatively-charged electrons embedded within a positively-charged "soup."

4 0
3 years ago
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A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is v  = 0.0002808 \ m/s

Explanation:

From the question we are told that

    The current on the copper is  I  = 20 \ A

     The cross-sectional area is  A =  5.261 \ mm^2 =  5.261 *10^{-6} \ m^2

The number of copper atom in the wire is  mathematically evaluated

      n  =  \frac{\rho *  N_a}Z}

Where \rho is the density of copper with a value \rho =  8.93 \ g/m^3

          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

     n  = 8.46 * 10^{28}  \  atoms /m^3

Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         N  = 8.46 * 10^{28}  \  electrons

The current through the wire is mathematically represented as

         I  =  N * e * v * A

substituting values

        20 =  8.46 *10^{28} * (1.60*10^{-19}) * v *  5.261 *10^{-6}

=>     v  = 0.0002808 \ m/s

       

8 0
3 years ago
Question 15 plz picture above
Pie
15) C. The amount of each element that begins....
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3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Alekssandra [29.7K]

so, question number 10 answer is 82 watts

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3 years ago
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