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EastWind [94]
3 years ago
8

What would earth be like without the moon?

Physics
2 answers:
sleet_krkn [62]3 years ago
8 0
<span>Earth without its moon would be a very different world indeed. No eclipses. Smaller tides. But the biggest change would be in the length of Earth’s day.</span>
maxonik [38]3 years ago
6 0
1.       Nights would be much, much darker. The next brightest object in the night sky is Venus. But it still wouldn't be enough to light up the sky. A full moon is nearly two thousand times brighter than Venus is at its brightest.

2.       Without the moon, a day on earth would only last six to twelve hours. There could be more than a thousand days in one year! That's because the Earth's rotation slows down over time thanks to the gravitational force -- or pull of the moon -- and without it, days would go by in a blink.

3.       A moonless earth would also change the size of ocean tides -- making them about one-third as high as they are now. 

4.       Forget about seeing any lunar eclipses -- or any solar eclipses -- without the moon, there would be nothing to block the sun.

5.       Without a moon the tilt of our earth's axis would vary over time. This could create some very wild weather. Right now, thanks to our moon, our axis stays tilted at twenty-three point five degrees. But without the moon the earth might tilt too far over or hardly tilt at all leading to no seasons or even extreme seasons. 

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A 0.5-kilogram ball is thrown vertically upward with an initial kinetic energy of 25 joules. Ap-proximately how high will the ba
tatuchka [14]
It's gravitational potential energy at the top will roughly equal it's kinetic energy when it was released (a little is lost to air resistance).  Note this will assume the release point is zero potential energy.  (we are free to define it that way, just letting you know).  Gravitational potential energy is mgh.
mgh=25J
h=25J/(0.5kg x 9.81m/s^2) = 5.097m
So it goes about 5.1 meters above the point where it was released
4 0
3 years ago
A 10.0-kg animal is running at a rate of 3.0 m/s. What is the kinetic energy of the animal?
Ilya [14]
Kinetic energy is calculated by using formula:

Ek = 1/2*m*v^2

Now we replace values of mass and speed in this formula.

Ek = 1/2*10*3^2 = 45.0J

When you have a formula and all variable values are given in the text of question just replace variable values in equation and find the final value.
4 0
3 years ago
A space-walking astronaut has become detached from her spaceship. She's floating in space while holding a wrench. She is thinkin
ser-zykov [4K]

Answer:

please find the solution which is defined as follows:

Explanation:

Throughout the opposite direction, she not able to throw her tool-belt. In this scenario, she will be sending her floating through her ship. Unless interrupted, it could refer to Newton's The rule of it in motion remains in motion. Consequently, if they throw it one way because there are no molecules to interrupt your course, you can continue to go the other way.

6 0
3 years ago
When a box of chocolate bars is in mechanical equilibrium, what can be correctly said about all the forces that act on it?
rjkz [21]

Answer:

All the forces are opposite and equal and will give net force zero.

Explanation:

Mechanical equilibrium is the equilibrium in which the total force on the system is zero means the system is neither accelerated nor any kind of torque on the system.

The mechanical system can also be defined as the equal forces are applied in opposite direction in a system which cancels out all the forces, will give net force zero.

Therefore, when a box of chocolate bars is in mechanical equilibrium all the forces in this system  are equal and the opposite which balances each other will give net force zero.

3 0
3 years ago
Saturated water vapor at 200 kPa is condensed into a saturated liquid via a constant-pressure process inside of a piston-cylinde
evablogger [386]

Answer:

The process is not possible

Explanation:

if we want to determine if the process is possible , we can check with the second law of thermodynamics

ΔS≥ ∫dQ/T

for a constant temperature process ( condensation)

ΔS≥ 1/T ∫dQ

and from the first law of thermodynamics

ΔH = Q - ∫VdP , but P=constant → dP=0 → ∫VdP=0

Q=ΔH

then

ΔS≥ΔH/T

from steam tables

at P= constant = 200 Kpa → T= 120°C = 393 K

at P= constant → H vapor = 2201.5 kJ/kg ,  H liquid = 1.5302 kJ/kg

, S vapor= 7.1269 kJ/kg , S liquid 1.7022 kJ/kg

therefore

ΔH = H vapor - H liquid = 2201.5 kJ/kg -  1.5302 kJ/kg = 2199.9698 kJ/kg

ΔS = S vapor - S liquid = 7.1269 kJ/kg - 1.7022 kJ/kg = 5.4247 kJ/kg

therefore since

ΔS required  = ΔH/T = 2199.9698 kJ/kg/(393 K)= 5.597 kJ/kg K

and

ΔS= 5.4247 kJ/kg  ≤ ΔS required=5.597 kJ/kg K

the process is not possible

5 0
3 years ago
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