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3 years ago

PLEASE HELP MEEEEEE!!!!!!!!!!!!!!!!!!

Physics

2 answers:

Simora [160]3 years ago

5 0

You would need to freeze it in a freezer. Hope this helps if it does could I have brainlist thanks

MAXImum [283]3 years ago

3 0

When liquid water loses thermal energy, it undergoes freezing : changing state from a liquid to a solid. We see many examples of this in everyday life. Puddles, ponds, lakes, and even parts of oceans freeze when the water becomes cold enough. At low temperatures, Earth's surface water freezes and forms solid ice.

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A driver brings a car to a full stop in 2.0 s. If the car was

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Working displayed in the picture below, the answer is -11 m s^-2

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3 years ago

Object 1 and 2 attract each other with a electrostatic force of 36.0 units. If the charge of object 1 is changed to four times t

Mrac [35]

The answer would be 6

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3 years ago

Suppose that instead of being inclined to Earth's orbit around the Sun, the Moonâs orbit was in the same plane as Earthâs orbit

ycow [4]

Answer:

c) 12

Explanation:

A Solar eclipse occurs when The Sun, The Earth and The Moon comes in a straight line with the Moon being in between the Earth and the Sun. At this point the Moon appears to block the Sun and Moon's shadow falls on Earth. This would occur only on the day of the New Moon.

If the Moon's orbit was in the same plane as that of the Earth's orbit. Every new Moon, there would be a Solar Eclipse. The Lunar cycle is of 29.5 Days which means there will be one new Moon every month. So there will be 12 Solar Eclipses every year.

Currently, the orbit of the Moon is tilted at an angle of 5° thus we don't see that many Solar eclipses. Maximum of 5 solar eclipses can occur in an year.

6 0

3 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

3 years ago

What is the resolution of three skeleton key?

Arisa [49]

It depends because it’s might be lolilolololol 21212132

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3 years ago