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jolli1 [7]
3 years ago
5

Can someone please write a summery about Yes day the movie from 2021 and it has to have something about Hygiene in it and this i

s for 100 points so can you please help me I will give you $100 and 100 points for a 2 page essay

Physics
2 answers:
qwelly [4]3 years ago
8 0

Answer:                           This book is about a mom that is always saying no to her kids and is always pushing them to focus on what they are doing, but one day the mom or dad runs into a man that does what he calls "A Yes Day" with his kids and what this is, is for 24 hours you say yes to whatever you want to do and you aren't able to say no. So the mom tries yes day with her kids and they go through many crazy/extreme events. This has to deal with hygiene because the kids trash the house and get very dirty in a few parts of the movie. They always try to be as clean as possible, but the limits have been broken... This is why I think this movie has to deal with hygiene.

ella [17]3 years ago
3 0

Answer:

I can't do 2 pages but I'll try

Explanation:

Yes day is a movie about parents and kids who usually fought, until someone suggested a yes day a day when parents can't say no. In the beginning of the movie, you'll first see that the parents used to say yes to everything, until they has kids they started saying no to everything. When someone suggested yes day they had to earn it, they did all the chores and they got a yes day. They made a list of 5 things they wanna do during yes day. i don't know what else to put for hygiene but yea.

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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.
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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

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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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