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3 years ago

In what states does water appear on earth and on other planets?

Physics

2 answers:

Marysya12 [62]3 years ago

7 0

Water appears in the liquid state on other planets.

aleksley [76]3 years ago

4 0

Water can exist as solid, liquid or gas on Earth and other planets. When water exists as a solid it is called ice; as a liquid it is called water and as a gas it is called water vapor. The state that the water exists depends on the temperature of the planet. In areas where the temperature is below 0 degrees Celsius, water would exist as a solid. Between 0 degrees and 100 degrees Celsius water will exist as a liquid. Water vapor is formed at temperatures higher than 100 degrees Celsius. Water vapor, however, can exist at temperatures below 100 degrees Celsius. This occurs when water evaporates. In this case, water molecules move out of the liquid state into the gaseous state due to their energy.

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A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

Lubov Fominskaja [6]

Answer:

The moment of inertia is $I=\frac{M}{12} a^{2}$

Explanation:

The moment of inertia is equal:

$I=\int\limits^a_b {r^{2} } \, dm$

If r is $-\frac{a}{2}$

and $dm=\frac{M}{a} dr$

$I=\int\limits^a_b {r^{2}\frac{M}{a} } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}$

$I=\frac{M}{a} \int\limits^a_b {r^{2} } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\ I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}$

7 0

3 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

7 0

3 years ago

Consider a household that uses 14,000 kWh of electricity and 900 gal of fuel oil, per year, during a heating season. The average

ryzh [129]

Answer:

9517.2 lbm

Explanation:

Electricity consumption = 14000 kWh/year

Fuel consumption = 900 gal/year

Amount of CO₂ produced per gallon = 26.4 lbm/gal

Amount of CO₂ produced per kWh = 1.54 lbm/kWh

Amount of CO₂ produced in one year

$14000\times 1.54+900\times 26.4=45320\ lbm$

Reduction would be

$0.21\times 45320=9517.2\ lbm$

The reduction in the amount of CO₂ produced is 9517.2 lbm

7 0

3 years ago

Two cars approach each other. How does the

Lemur [1.5K]

Answer:

<u>The car's fast. The ground isn't moving.</u>

Hope this helped! :D

7 0

2 years ago

What type of energy transformation is taking place when natural gas is used to heat water

bazaltina [42]

Thermal energy transformation is taking place.

3 0

3 years ago