why does diving 30m below sea level affect our bodies more than being in a building 30m above sea level

1 answer:

5 0

Imagine you are in a swimming pool 30m deep. Assuming you know that water is denser than air, you would know that the 30m of water above you will carry more weight, and press down on your body. Say you were in a swimming pool 60m deep, you would be sandwiched between 30m of water pressing down on you, and the upthrust created by the 30m of water below you.

In a building 30m up, the pressure will be regulated, as you are in a building. The floor will be strong enough to support the weight of the body, and the body will not recoil into itself.

You might be interested in

50 POINTS!!!

natulia [17]

Answer: In a divergent plate boundary, seafloor spreading taking place. It leads to the formation of oceans as new materials are added here along the mid-oceanic ridge. There occur volcanism and shallow-focus earthquakes.

In a convergent plate boundary, two plates collide to form high mountain belts and also volcanic eruptions take place. There occur long chains of volcanic as well as island arcs, in association with deep-focus earthquakes.

In a transform plate boundary, two plates slide past each other, conserving the plates. Shallow-focus earthquakes are generated here.

The earth has experienced various geological processes, such as weathering and erosion of rocks, earthquakes, volcanic eruptions, mass extinction events, plate tectonic movements and many more. These continuous processes have configured the present shape of the earth's surface.

For example, the breaking up of the supercontinent Pangea divided into Laurasia and Gondwanaland and subsequently formed the present scenario. This separation of continents has taken place due to the convection current that generates in the mantle.

5 0

3 years ago

Read 2 more answers

1. What is true about all forces?a. They are unbalanced b. They involve more than one object c. They cause objects to moved. d.

Pachacha [2.7K]

the answer is most likely D

3 0

3 years ago

A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

shusha [124]

Answer:

$v=d\sqrt{\frac{k}{m}}$

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

$U_{0}+K_{0}=U_{f}+K_{f}$

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

$U_{0}=K_{f}$