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xenn [34]
4 years ago
6

How could the absence of gravity affect the formation of planets

Physics
1 answer:
Airida [17]4 years ago
8 0
It would be very chaotic... For example anything that was not attached to the Earth will probably go flying off into space in the straight linear line. if you're already disappeared the Earth's atmosphere would float in space so other planet's atmospheres would also float into space... planets will break apart into pieces and go off in space and planets are held together by gravity period. Imagine gravity is a string, once you look over that string or quote unquote "turn off" gravity, everything will go flying into linear lines on to the vastness of the cosmos. I hope this helps if you need any more examples for any other questions please let me know have a blessed day. :-)
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A temperature of 20°C is equal to ? °F.
Papessa [141]
The answer is 68 F. i hope this helps
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4 years ago
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What is the most likely outcome of increasing the number of slits per unit distance on a diffraction grating?1) lines become nar
Nina [5.8K]

It's important to know that diffraction gratings can be identified by the number of lines they have per centimeter. Often, more lines per centimeter is more useful because the images separation is greater when this happens. That is, the distance between lines increases.

<h2>Therefore, the answer is 2.</h2>
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2 years ago
Which means for obtaining hydrogen from water would require the most energy??
Flauer [41]
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8 0
3 years ago
What is the speed of a car that traveled 500 meter in 30 seconds?
GenaCL600 [577]
<h2>Greetings!</h2>

To find speed, you need to remember the formula:

Speed = distance ÷ time

So plug the given values in:

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3 0
3 years ago
Calculate the heat, in kilocalories, that is absorbed if 183 g of ice at 0.0 ∘C is placed in an ice bag, melts, and warms to bod
boyakko [2]

Answer:

The total amount of heat needed will be Q_T=21.411kcal.

Explanation:

We will divide the calculation in two: First, the heat needed to melt the ice, and then the heat needed to warm the resulting liquid from 0°C to 37°C.

m=183g

l_f=80\frac{cal}{g} =334\frac{J}{g}

l_w=1\frac{cal}{g} =4.184\frac{J}{g}

<em>i) </em>The fusion heat will be:

Q_f=l_fm=14640cal=14.640kcal

<em>ii)</em> The heat needed to warm the water from T_i=0^{\circ}C to T_i=37^{\circ}C will be:

Q_w=l_wm(T_f-T_i)=6771cal=6.771kcal

So, the total amount needed will be the sum of these two results:

Q_T=Q_f+Q_w=14.640kcal+6.771kcal=21.411kcal.

8 0
3 years ago
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