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Mrrafil [7]
2 years ago
12

Describe how reactivity changes as you go down Group 1A.

Physics
1 answer:
kenny6666 [7]2 years ago
3 0

Answer:

it is a the answer is a btw

Explanation:

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Answer:

1.The calorie was originally defined as the amount of heat required at a pressure of 1 standard atmosphere to raise the temperature of 1 gram of water 1° Celsius. Since 1925 this calorie has been defined in terms of the joule, the definition since 1948 being that one calorie is equal to approximately 4.2 joules.

2.Boiling water at 100 degrees Celsius: 540 calories are needed to turn 1 gram (at 100 degrees Celsius) of water to steam.

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Why does fluid pressure exist?
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Fluid pressure from gravity is the weight of the fluid above divided by the area it is pushing on. Fluid pressure applies in all directions. Internal pressure of an object equals the external fluid pressure, otherwise the object could be crushed. Wind and heating can also create pressure.

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During crystallisation the crystals separate out from the hot ________solution of a substance on cooling
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The process of separation or deposition of crystals from a hot saturated solution on gentle cooling of the solution is called 'crystallisation'.

Explanation:

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2 years ago
What is newton's 3rd law of physics ​
ValentinkaMS [17]

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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.

3 0
3 years ago
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