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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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How is the pressure of a gas related to its concentration of particles?A) Pressure will expand a gas, enlarging its volume and r

Lesechka [4]

Answer:

C) Pressure will compress a gas, reducing its volume and giving it a greater density and concentration of particles.

Explanation:

At constant temperature, pressure and volume are inversely related.

P V = constant

$\Rightarrow P \propto \frac{1}{V}$

As the pressure increases, the gas compresses, the particles come closer reducing the volume of gas.

As we know, with decrease in volume, density increases.

$Density = \frac{Mass}{Volume}$

$Density \propto \frac{1}{Volume}$

Thus, the pressure of a gas is directly related to concentration of particles. Increase in pressure causes increase in concentration of the particles.

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3 years ago

What is the magnification of a virtual image if the image is 60.0 cm from a

Hatshy [7]

Answer:

4cm

Explanation:

Magnification of the virtual image

= image distance / object distance

Given

Image distance = 60.0cm

Object distance = 15.0cm

Therefore,

Magnification = 60.0/15.0

= 4 cm

6 0

3 years ago

Read 2 more answers

One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side

anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

$L1^2=L2^2=L^2=2^2+(H/2)^2$ Replacing this value in the previous equation:

$\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34$ Solving for H:

$H=\sqrt{52}$

We can now, calculate the angle between L1 and the 2m segment:

$\alpha = atan(\frac{H/2}{2})=60.98°$

If we make a sum of forces in the midpoint of the rope we get:

$-2*T*cos(\alpha ) + F = 0$ where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

$T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N$

7 0

3 years ago

Kinetic and

beks73 [17]

Answer:

raise the board to a higher angle

Explanation:

Static friction is the force opposite to the applied force.

Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.

So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.

Hence, the correct option is "raise the board to a higher angle".

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3 years ago

6. What's a stalactite?

iogann1982 [59]

I think it's d......... Good luck

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3 years ago