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3 years ago

The temperature scale which starts at absolute zero is the _____.

2 answers:

6 0

The temperature scale which starts at absolute zero is the Kelvin scale. The correct option in respect to the given question is the last option. William Thompson was the British scientist and inventor that invented the Kelvin scale. William Thompson was also popularly known as Lord Kelvin.His discovery of the Kelvin scale is considered one among the three best scales in use for measuring temperatures.Each measuring unit of this scale is never called a degree but a Kelvin. This specialized scale gives the option of measuring temperature in both centigrade and Fahrenheit.

Stella [2.4K]3 years ago

6 0

The temperature that the scale starts at absolute zero is called the Kelvin scale

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so you can see all the different categories at once. both as a whole and on an individual scale.

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What happens to the magnitude of the force of gravitation between two objects if: 1. distance between the objects is tripled? 2.

Ludmilka [50]

<span>2. mass of both objects doubled?
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3 years ago

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Brums [2.3K]

Answer: two waves with identical crests and troughs meet

Explanation:

My teacher gave me the answer

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3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

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3 years ago

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melisa1 [442]

Answer:

1.d

2.d

3.c

4.c

Okay, d & d, c & c. Iknow what you're thinking, I guessed. Nope! I just took a good look and chosed what I thought was the best answer

Explanation:

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3 years ago