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vlada-n [284]
3 years ago
9

What is the relationship between atmospheric pressure and the density of gas particles in an area of decreasing pressure? (2 poi

nts)
A. As air pressure in an area decreases, the density of the gas particles in that area decreases.
B. As air pressure in an area decreases, the density of the gas particles in that area increases.
C. As air pressure in an area decreases, the density of the gas particles in that area remains constant.
D. As air pressure in an area decreases, the density of the gas particles in that area increases and decreases in an alternating pattern.

Also it's not showing earth space science, so that's why I put it as Physics.
Physics
2 answers:
Effectus [21]3 years ago
8 0

Answer:

As air pressure in an area decreases, the density of the gas particles in that area decreases.

Explanation:

ivann1987 [24]3 years ago
4 0

Answer:

A. As air pressure in an area decreases, the density of the gas particles in that area decreases.

Explanation:

At constant temperature, pressure and density of a gas are directly related.  So if pressure decreases, density decreases.

You might be interested in
Select the correct answer.
Dahasolnce [82]

Answer:

a is the correct choice

Explanation:

5 0
3 years ago
Read 2 more answers
What happens to the kinetic energy of a body when: a) the mass of the body is doubled at constant velocity? b) the velocity of t
blagie [28]
Using the formula KE=1/2mv^2

a: The kinetic energy doubles.
b: The kinetic energy quadruples.
c: The kinetic energy is cut in half.
Hopefully it’s clear how the formula can show you this.
3 0
3 years ago
A uniform electric field has a magnitude 1.80 kV/m and points in the +x direction. (a) What is the electric potential difference
EastWind [94]

Answer:

a)ΔV = 6.48 KV

b)ΔU =18.79 mJ

Explanation:

Given that

E= 1.8 KV/m

a)

We know that

Electric potential difference  ΔV given as

ΔV = E .d

Here

E= 1.8 KV/m

d= 3.6 m

ΔV = E .d

ΔV = 1.8 x 3.6 KV

ΔV = 6.48 KV

b)

Given that

q=+2.90 µC

Change in electric potential energy ΔU given as

ΔU = q .ΔV

\Delta U=2.9\times 10^{-6}\times 6.48\times 10^3\ J

ΔU =18.79 mJ

8 0
3 years ago
A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is
blagie [28]

Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

The total normal force acting = normal reaction + 24N

Note that the normal reaction is always equal the weight of the table

Hence the normal force acting in the block is 244.0+24 = 268.0N

4 0
3 years ago
To move a heavy object, like a refrigerator what could be used to help decrease the frictional force?
Olenka [21]

Answer:

A. Remove everything in the refrigerator to lighten the load.

B. Put a lubricant between the surface of the object and the floor

C. Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily

Explanation:

Force of friction is a resistance force which acts between two surfaces which are in relative motion. Friction is both boon and bane. Due to friction, we are able to sit, walk etc but also, due to friction there is dissipation of energy. Friction can be reduced by applying lubricants, reducing contact area, reducing the load.

F = μN where N is the normal force which depends on the mass.

Thus, by reducing the load, force of friction can be reduced. Round objects like wheels can also be used. By this the contact area reduces.

8 0
3 years ago
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