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3 years ago

What characteristic distinguishes a group from an aggregate?

Physics

2 answers:

Elis [28]3 years ago

8 0

Explanation:

Hello!

Let's solve this!

Next we will see the differences between group and aggregate:

A group of people are people who are together and who also have something in common or carry out an activity in common.

The main characteristic of a group is the interaction between its members.

In the case of the aggregate, they are people who are in the same place but do not share anything.

docker41 [41]3 years ago

3 0

Group is a collection of people that are in the same place and share some characteristics. examples are a group of students, a group of engineers and etc. while aggregate is a collection of people that are in the same place but do not share anything in common

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g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7

bogdanovich [222]

Answer:

When they are approaching each other

$f_a = 2228.7 \ Hz$

When they are passing each other

$f_a = 2100Hz$

When they are retreating from each other

$f_a = 1980.7 Hz$

Explanation:

From the question we are told that

The velocity of car one is $v_1 = 13.0 m/s$

The velocity of car two is $v_2 = 7.22 m/s$

The frequency of sound from car one is $f_e = 2.10 kHz$

Generally the speed of sound at normal temperature is $v = 343 m/s$

Now as the cars move relative to each other doppler effect is created and this can be represented mathematically as

$f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ]$

Where $v_s$ is the velocity of the source of sound

$v_o$ is the velocity of the observer of the sound

$f_o$ is the actual frequence

$f_a$ is the apparent frequency

Considering the case when they are approaching each other

$f_a = f_o [\frac{v + v_o}{v - v_s} ]$

$v_o = v_2$

$v_s = v_1$

$f_o = f_e$

Substituting value

$f_a = 2100 [\frac{343 + 7.22}{ 343 - 13} ]$

$f_a = 2228.7 \ Hz$

Considering the case when they are passing each other

At that instant

$v_o = v_s = 0m/s$

$f_o = f_e$

$f_a = f_o [\frac{v }{v } ]$

$f_a = f_o$

Substituting value

$f_a = 2100Hz$

Considering the case when they are retreating from each other

$f_a = f_o [\frac{v - v_o}{v + v_s} ]$

$v_o = v_2$

$v_s = v_1$

$f_o = f_e$

Substituting value

$f_a = 2100 [\frac{343 - 7.22}{343 + 13} ]$

$f_a = 1980.7 Hz$

7 0

3 years ago

Sand dunes during the Dust Bowl are an example of:

choli [55]

Sand dunes during the Dust Bowl are an example of windward.

The sand dunes are formed due to high movement of wind in the deserts which make dunes in the direction where wind is moving. Windward refers to facing the wind or situated on the side facing the wind so in those regions where the wind moves bring sand with itself and make dunes in that area so we can conclude that sand dunes are formed in the windward regions.

Learn more: brainly.com/question/24879138

5 0

3 years ago

The decimal equivalent for meter is

USPshnik [31]

m =dm ______ 10.000

Meters

The metre is a unit of length in the metric system, and is the base unit of length in the International System of Units (SI).

As the base unit of length in the SI and other m.k.s. systems (based around metres, kilograms and seconds) the metres is used to help derive other units of measurement such as the newton, for force.

8 0

3 years ago

When you are using a crowbar to lift a large rock, you are working agents the force called

TEA [102]

When you're using a crowbar to lift a large rock, you are working against the force called

$\boxed {\bf Gravity}$

$\bf Explanation:$
Gravity on Earth is what gives weight to all objects, it's defined as all things that have mass or energy are gravitated towards each other. Therefore when you're using a crowbar to lift a large rock, the weight is caused by gravity.

I hope this helps you!

7 0

3 years ago

A person throws a ball straight up in the air. The ball rises to a maximum height and then falls back down so that the person ca

Lana71 [14]

Answer:

The acceleration is about 9.8 m/s2 (down) when the ball is falling.

Explanation:

The ball at maximum height has velocity zero

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s² (positive downward and negative upward)

$v=u+at\\\Rightarrow 0=u-9.8\times t\\\Rightarrow u=9.8t$

The accleration 9.8 m/s² will always be acting on the body in opposite direction when the body is going up and in the same direction when the body is going down. The acceleration on the body will never be zero

5 0

3 years ago