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GarryVolchara [31]
3 years ago
9

What is the relationship between the intervals on the Celsius scale and the Kelvin scale

Physics
1 answer:
MAXImum [283]3 years ago
4 0

An interval of 1 degree Celsius is equal to 1 Kelvin

Explanation:

There are different scales used for measuring temperatures. The most common ones are:

- Celsius scale: it is set by the following convention: the zero in this scale is placed at the freezing temperature of the water, while the 100 degrees is placed in corresponding to the boiling point of water. The the scale is divided into 100 equal intervals between these two points. Temperatures in this scale are indicated with the Celsius degree (^\circ{}C.

- Kelvin scale: it is set such that the absolute zero (the temperature at which particles in matter stop to have any motion) is placed at zero in this scale. A difference of 1 degree in this scale is equal to 1 degree Celsius. Temperatures in this scale are indicated with the Kelvin (K).

The conversion between the two temperatures is:

T(^{\circ}) = T(K) - 273.15

So for example, the absolute zero (0 K) corresponds to -273.15^{\circ}C in the Celsius scale. However, intervals between two temperatures are measured in the same way in the two scales: therefore, an interval of 1 degree in the Celsius scale is equal to an interval of 1 Kelvin in the Kelvin scale.

Learn more about temperatures here:

brainly.com/question/1603430

brainly.com/question/4889772

#LearnwithBrainly

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You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone
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Answer:

0.37 m

Explanation:

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$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$

$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$

$v= 2.69$ m/s

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