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3 years ago

Increasing air pressure indicates which of the following types of air is moving into your area?

Physics

2 answers:

NISA [10]3 years ago

5 0

<h3>Answer;</h3>

A. cold, dry air

Increasing air pressure indicates that cold, dry air is moving into your area.

<h3>Explanation;</h3>

Cold temperatures can create areas of increased air pressure because cold air has a higher density and the concentration of molecules can cause an increase in air pressure.
The change in pressure is caused by the air density which is related to temperature. Warm air is less dense than cold air since the gas molecules in warm air travel faster and are farther apart. Therefore, and are with high pressure has a dense air mass where the temperature is cool.

Yakvenalex [24]3 years ago

4 0

For the answer to the question above, it is multiple choice letter a. Cold, dry air.
You must know that the pressure changes and the patterns of circulation of the upper air.

I hope my answer helped you. have a nice day!

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3 years ago

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

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a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

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Package is 44.1 meter below helicopter

Explanation:

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

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b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

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v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

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$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

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