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Sauron [17]
4 years ago
6

Why is the definition for second based on the Cesium 133 atom?

Engineering
1 answer:
jarptica [38.1K]4 years ago
7 0

Answer:

  The rate of vibrations of the atoms is consistent

Explanation:

A second is defined as 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.

The definition applies when the atom is in isolation at a temperature of absolute zero. Adjustments must be made for the conditions that can be maintained in practice.

Such radiation is determined by the laws of quantum mechanics, so is extremely consistent.

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HW6P2 (20 points) The recorded daily temperature (°F) in New York City and in Denver, Colorado during the month of January 2014
Maurinko [17]

Answer & Explanation:

function Temprature

NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];

DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];

%AVERAGE CALCULATION AND ROUND TO NEAREST INT

avgNYC=round(mean(NYC));

avgDEN=round(mean(DEN));

fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);

fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);

%part B

count=1;

NNYC=0;

NDEN=0;

while count<=length(NYC)

   if NYC(count)>avgNYC

       NNYC=NNYC+1;

   end

   if DEN(count)>avgDEN

        NDEN=NDEN+1;

   end

   count=count+1;

end

fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);

fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);

%part C

count=1;

highDen=0;

while count<=length(NYC)

   if NYC(count)>DEN(count)

       highDen=highDen+1;

   end

   count=count+1;

end

fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);

end

%output

check the attachment for additional Information

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4 years ago
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Answer:

E = 2940 J

Explanation:

It is given that,

Mass, m = 12 kg

Position at which the object is placed, h = 25 m

We need to find the potential energy of the mass. It is given by the formula as follows :

E = mgh

g is acceleration due to gravity

E=12\times 9.8\times 25\\\\E=2940\ J

So, the potential energy of the mass is 2940 J.

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Answer:

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Explanation:

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