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 swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$