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3 years ago

The force of air particles over an area is?

Physics

2 answers:

Arada [10]3 years ago

4 0

That's the "air pressure" everywhere on that area.

alex41 [277]3 years ago

4 0

<h3><u>Answer;</u></h3>

Air pressure

<h3><u>Explanation</u>;</h3>

<em><u>Pressure is the force that is applied over a unit of area. Air pressure is the weight of air molecules pressing down on the Earth. It is the force exerted by air particles on the earth's surface.</u></em>
The column of air extends to the top of the atmosphere. As one move up increasing the altitude, the air pressure decreases since the air column is reduced.

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A spring scale hung from the ceiling stretches by 5.2 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl

pantera1 [17]

Physics - Damon, Wednesday, December 9, 2015 at 5:13am
F = k x

k = 2 g/6.1 cm

2.5g = (2g/6.1cm) x

x = 6.1 (2.5/2) cm

8 0

3 years ago

Read 2 more answers

the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m

olga_2 [115]

Answer:

Spring's displacement, x = -0.04 meters.

Explanation:

Let the spring's displacement be x.

Given the following data;

Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg

Number of shrews, n = 49

Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

Fnet = Felastic + Fg = 0

But, Felastic = -kx

Total mass, Mt = nm

Fg = -Mt = -nmg

-kx -nmg = 0

Rearranging, we have;

kx = -nmg

Making x the subject of formula, we have;

$x = \frac {-nmg}{k}$

Substituting into the formula, we have;

$x = \frac {-49*0.002*9.8}{24}$

$x = \frac {-0.9604}{24}$

x = -0.04 m

Therefore, the spring's displacement is -0.04 meters.

3 0

2 years ago

For a moving object, the force acting on the object varies directly with the object's acceleration. When a force of 81 N acts on

tensa zangetsu [6.8K]

Answer:

18 N

Explanation:

You have to do:

81 N: 9 ms2= x: 2 ms2

6 0

3 years ago

Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass

Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

$\Delta W = \Delta PE + \Delta KE$

Where,

PE = Potential Energy

KE = Kinetic Energy

$\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2$

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

$P = \frac{\Delta W}{\Delta t}$

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

The rate of mass flow is,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

Where,

$\rho_w$ = Density of water

A = Area of the hose $\rightarrow A=\pi r^2$

The given radius is 0.83cm or $0.83 * 10^{-2}$ m, so the Area would be

$A = \pi (0.83*10^{-2})^2$

$A = 0.0002164m^2$

We have then that,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

$\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)$

$\frac{\Delta m}{\Delta t} = 1.16856kg/s$

Final the power of the pump would be,

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

$P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)$

$P = 57.1192W$

Therefore the power of the pump is 57.11W

6 0

3 years ago

Jeremiah is conducting an investigation about the water cycle. He is given the following

Vlada [557]

The water cycle outlines the continuous water movement in liquid, solid and gaseous state between locations on the Earth's surface.

The glass jar represents the lake while the atmosphere is represented by the space above the water, and the sky is represented by the (clear) plastic wrap

Arrangement description and Processes;

The processes of the water cycle includes;

Evaporation;

Condensation

Precipitation

Sublimation

Runoff

Infiltration

The arrangement of the materials is as follows;

Place the glass jar (the lake) containing water and the lamp (the Sun) side by side, such that the lamp light shines on the water surface

Cover the glass jar by wrapping the plastic wrap (the sky) around it to prevent the escape of water vapor when the water is hot.

Switch on the lamp so that it heats the water by radiation heat transfer

Observed processes;

The processes demonstrated by the above experiment includes;

1) Evaporation: As the water in the glass jar becomes warmer, the level of the water in the jar can be observed to decrease slightly due to evaporation

2) Condensation: Fog formation, Clouds

When hotter, the water surface as seen through the clear plastic wrap becomes less clearer due to evaporation, and condensation of the vapor while floating above the water surface, similar to the clouds seen in the sky.

3. Precipitation: Rain;

The clear plastic wrap covering the top of the glass jar, prevents the movement of the vapor further away, such that the tiny condensed vapor gather together, to form big droplets under the plastic wrap that falls back into the jar, which is similar to the process of rainfall

The above processes are repeated as more water evaporates from the jar condenses on the plastic wrap and falls back into the jar, showing the process by which water is recycled from the lake into the atmosphere and back to the lake.

Learn more here:

brainly.com/question/2430469

4 0

2 years ago