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3 years ago

Why does pressure change with elevation and depth?

2 answers:

5 0

<span>For any "block" of fluid, the pressure at the bottom times the area must equal the weight plus the pressure at the area times the area, if equilibrium is to be maintained. If not, the "block" of air (some of it at least) will move down or up. Therefore the pressure at the bottom is always higher.

Hope this helps :)
</span>

nasty-shy [4]3 years ago

3 0

<span>the deeper you go the more pressure...that's for depth</span>

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The power radiated by the sun is 3.90 1026 W. The earth orbits the sun in a nearly circular orbit of radius 1.50 1011 m. The ear

WITCHER [35]

Answer:

3234.2 W

Explanation:

Since intensity I = Power/Area. The intensity of the light from the sun, I = power radiated by sun/area of sphere of radius, r = 1.5 × 10¹¹ m.

So, I = 3.9 10²⁶W/4π(1.5 × 10¹¹ m)² = 2.069 × 10³ W/m².

Now, the power radiated on the patch of area 0.570 m² at the equator is

P = Icos27/A = 2.069 × 10³ W/m² cos27/0.570 m² = 1843.49/0.570 = 3234.2 W

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3 years ago

In one experiment, the students allow the block to oscillate after stretching the spring a distance A. If the potential energy s

atroni [7]

Answer:

K = m g (A - A2)

Explanation:

In a block spring system the total energy is the sum of the potential energy plus the kinetic energy, for maximum elongation all the energy is potential

Em = U₀ = m g A

For when the system is at an ele

Elongation A2 less than A, energy has two parts

Em = K + U₂

K = Em –U₂

We substitute

K = m g A - m gA2

K = m g (A - A2)

3 0

4 years ago

Read 2 more answers

The diagram below shows the velocity vectors for two cars that are moving relative to each other.

scoundrel [369]

Answer:

The answer is " $5 \ \frac{m}{s} \ west$ "

Explanation:

$\to \vec{V_1} = (25 \frac{m}{s}) (\hat{-i})\\\\\to \vec{V_2} = (20 \frac{m}{s}) (\hat{-i})\\\\$

velocity of car | respect to car :

$\to \vec{V_{12}} = \vec{V_1} - \vec{V_2}\\\\$

$=\vec{-25} \hat{i}+ \vec{20} \hat{i}\\\\= 5 \ \frac{m}{s} \ west$

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3 years ago

A 10 kg box slides horizontally without friction at a speed of 1 m/s. At one point, a constant force is applied to the box in th

GrogVix [38]

Answer:

Fapp = 3 N

Explanation:

Applying the work-energy theorem, we know that the net work done on one object by an applied force, is equal to the change in the kinetic energy of the object.
This work is just the product of the applied force times the displacement, as follows:

$W_{net} = F_{app} * d (1)$