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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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Given:A=6x-2y B:-4x-8y C:-3x+9y. Commute A+B-C

DedPeter [7]

A+B-C
A = 6x - 2y
B = -4x - 8y
C = -3x + 9y

(6x - 2y) + (-4x - 8y) - (-3x + 9y)
(6x - 2y) + (-4x - 8y) + (3x - 9y)
2x -10y + (3x - 9y)

5x - 19y

8 0

3 years ago

If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ∘C , what is the difference between the ideal pressure (as predict

Rudik [331]

Answer:

1.98 atm

Explanation:

Given that:

Temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28 + 273.15) K = 301.15 K

n = 1

V = 0.500 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

P × 0.500 L = 1 ×0.0821 L atm/ K mol × 301.15 K

⇒P (ideal) = 49.45 atm

Using Van der Waal's equation

$\left(P+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$

R = 0.0821 L atm/ K mol

Where, a and b are constants.

For Ar, given that:

So, a = 1.345 atm L² / mol²

b = 0.03219 L / mol

So,

$\left(P+\frac{1.345\times \:1^2}{0.500^2}\right)\left(0.500-1\times 0.03219\right)=1\times 0.0821\times 301.15$

$P+\frac{1.345}{0.25}=\frac{24.724415}{0.46781}$

$P=\frac{24.724415}{0.46781}-\frac{1.345}{0.25}$

⇒P (real) = 47.47 atm

Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm

4 0

3 years ago

If a satellite weighs 321 lb. on the earth's surface (R = 4,000 miles), how much does it weigh 12,000 miles above the surface? (

Sever21 [200]

The gravitational force between the Earth and the satellite (its "weight") is inversely proportional to the distance between the centers of both objects.

On the surface, their centers are separated by 1 Earth radius.

12,000 miles above the surface, they're separated by 4 Earth radiii.

(4/1) = 4

So after the move, the satellite's weight is (1/4²) = 1/16 of its surface weight.

(321 lb) / (16) = (20 and a hair) lb

The correct choice from the given list is " >20 lb " .

3 0

3 years ago

Find expressions for the force needed to bring an object of mass m from rest to speed v in time t. express your answer in terms

VARVARA [1.3K]

Good morning.

We have that:

$\mathsf{V = a\cdot t}$ , since we have rest in the inicial time.

The acceleration can be found with Newton's Law:

$\mathsf{F = m\cdot a\iff a = \dfrac{F}{m}}$

Now we put the acceleratin in the velocity equation:

$\mathsf{V = \dfrac{F}{m} \cdot t}$

We want the force, so, let's isolate F:

$\mathsf{V\cdot m = F\cdot t}\\ \\ \\ \boxed{\mathsf{F = \dfrac{V\cdot m}{t}}}$

3 0

3 years ago

A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body of the top. The string has

AlladinOne [14]

Given Information:

Angular displacement = θ = 51 cm = 0.51 m

Radius = 1.8 cm = 0.018 m

Initial angular velocity = ω₁ = 0 m/s

Angular acceleration = α = 10 rad/s ²

Required Information:

Final angular velocity = ω₂ = ?

Answer:

Final angular velocity = ω₂ = 21.6 rad/s

Explanation:

We know from the equations of kinematics,

ω₂² = ω₁² + 2αθ

Where ω₁ is the initial angular velocity that is zero since the toy was initially at rest, α is angular acceleration and θ is angular displacement.

ω₂² = (0)² + 2αθ

ω₂² = 2αθ

ω₂ = √(2αθ)

We know that the relation between angular displacement and arc length is given by

s = rθ

θ = s/r

θ = 0.51/0.018

θ = 23.33 radians

finally, final angular velocity is

ω₂ = √(2αθ)

ω₂ = √(2*10*23.33)

ω₂ = 21.6 rad/s

Therefore, the top will be rotating at 21.6 rad/s when the string is completely unwound.

3 0

3 years ago