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Valentin [98]
2 years ago
9

According to max weber, flaunting of one’s wealth to show one’s status is called?

Physics
1 answer:
myrzilka [38]2 years ago
5 0

Status refers to a person's prestige, social honor, or reputation in society. Weber stated that political strength become now not rooted entirely in capital fees, but additionally in a single's individual popularity. Poets or saints, for example, can own a massive impact on society, often with little monetary worth.

The German sociologist Max Weber formulated a three-component theory of stratification that defines a status group (additionally status class and status estate) as a group of humans within a society who can be differentiated by means of non-financial characteristics along with honor, prestige, ethnicity, race, and religion.

Social stratification entails society as a system of hierarchical categories. Max Weber defined stratification because the division of a society into distinct communities, that have varying assignments of “status honor” or status.

Learn more about Max Weber here: brainly.com/question/13927529

#SPJ4

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A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m
elena-14-01-66 [18.8K]

Answer:

  • <u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

   f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration

<u>2. Formulae:</u>

         f=\dfrac{number\text{ }of\text{ }revolutions}{time}

          \omega=2\pi f

          a_c=\omega^2 r

           F_c=m\times a_c

<u>3. Solution (calculations)</u>

       f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}

       \omega=2\pi\times0.\overline{60}\approx 3.808rad/s

      a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2

      F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N

3 0
3 years ago
A rocket takes off from Earth. It travels 825km in 75 seconds. What is the
Naya [18.7K]

Answer:

11 km/s

Explanation:

v=s/t

v=825km/75s

v=11km/s

4 0
4 years ago
What is the relationship between area, force, and pressure? what is the relationship between area, force, and pressure? pressure
White raven [17]
Mathematically, relation between force, area and pressure is given by...
Pressure = force / area


hence, pressure is directly proportional to force but inversely proportional to area.
7 0
3 years ago
A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o
Jet001 [13]

Answer:

C) 3,000 kg m/s

Explanation:

We can consider the horizontal velocity of the motorcycle to be zero, since it rolls off the edge of the cliff very slowly. So, we only need to find the vertical velocity at the time of the impact with the ground.

The vertical velocity of the motorcycle at time t is given by (free-fall motion):

v(t)=v_0 -gt

where

v_0=0 is the initial vertical velocity (zero, since the motorcycle is not moving)

g = 9.8 m/s^2 is the acceleration due to gravity

t is the time

Since the motorcycle hits the ground after t = 3 seconds, we have

v(3 s)=0-(9.8 m/s^2)(3 s)=-29.4 m/s

And since we know its mass, m=100 kg, we can find its momentum:

p=mv=(100 kg)(-29.4 m/s)=-2940 kg m/s \cdot -3000 kg m/s

and the negative sign simply means downward direction.

8 0
3 years ago
A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.
zzz [600]

Answer:

1000 Hz

Explanation:

<em>The frequency would be 1000 Hz.</em>

The frequency, wavelength, and speed of a wave are related by the equation:

<em>v = fλ ..................(1)</em>

where v = speed of the wave, f = frequency of the wave, and λ = wavelength of the wave.

Making f the subject of the formula:

<em>f = v/λ.........................(2)</em>

Also, speed (v) = distance/time.

From the question, distance = 900 m, time = 3.0 s

Hence, v = 900/3.0 = 300 m/s

Substitute v = 300 and λ = 0.3  into equation (2):

f = 300/0.3 = 1000 Hz

6 0
3 years ago
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