Answer:
Because of gender, caste, race, wealth, and religion etc.
Explanation:
Social stratification means society is divided in different categories, class, layers or groups due to gender, caste, race, wealth, and religion etc.
Society stratifies due to the following regions:
(1) Gender discrimination means male- female difference.
(2) Unequal distribution of income and wealth
(3) Different types of religions
(4) Racism
(5) Type of education
(6) Social status etc.
Societies stratify, or divide their members into distinct groups or layers, based on various factors such as wealth, income, cultural beliefs, and status. Factors like prestige or age are also influential in some societies. Stratification systems can be either closed, allowing little social mobility, or open, where movement between classes is possible.
Societies stratify, or categorize people into different social standings, for various reasons. In many societies, stratification is an economic system, predominantly determined by wealth and income. Often, people interact chiefly with others of the same social standing, allowing economic and cultural factors to organize individuals into distinct groups or layers.
Societal stratification can also be driven by cultural beliefs that place value on specific attributes or characteristics such as prestige or age. For example, in some cultures, the elderly are esteemed, while in other societies, they are overlooked. Such cultural attitudes play a significant role in reinforcing stratification systems.
Also, stratification occurs when there is a difference in status or power between various societal roles, leading to a hierarchical organization of different groups - an example is the clear socioeconomic status (SES) division within society where individuals with more resources are seen at the top layer.
Closed and open stratification systems present themselves in different societies. Closed systems offer little opportunity for change in social position, whereas open systems, like class systems, are based on achievement, allowing movement and interaction between layers and classes.
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In order to solve this problem it is necessary to apply the concepts related to Newton's second law and the respective representation of the Forces in their vector components.
The horizontal component of this force is given as
F_x = Fcos(6.7)
While the vertical component of this force would be
F_y = Fsin(6.7)
In the vertical component, the sum of Force indicates that:
The Normal Force would therefore be equivalent to the weight and vertical component of the applied force, therefore:
In the horizontal component we have that the Force of tension in its horizontal component is equivalent to the Force of friction:
Using the previously found expression of the Normal Force and replacing it we have to,
Replacing,
Finally the acceleration would be by Newton's second law:
Therefore the greatest acceleration the man can give the airplane is
Answer:
The correct answer is d. tension pneumothorax.
Explanation:
The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.
2 A. What is the resistance of the iron?
3. A current of 0.2 A flows through an electric bell having a resistance of 65 ohms. What must be
the applied voltage in the circuit?
Answer:
(1) 0.04 ohms (2) 55 ohms (3) 13 volt
Explanation:
(1) The resistance of an electric device is 40,000 microhms.
We need to convert it into ohms.
To covert 40,000 microhms to ohms, multiply 40,000 and 10⁻⁶ as follows :
(2) Voltage used, V = 110 V
Current, I = 2 A
We need to find the resistance of the iron. Using Ohms law to find it as follows :
V = IR, where R is resistance
(3) Current, I = 0.2 A
Resistance, R = 65 ohms
We need to find the applied voltage in the circuit. Using Ohms law to find it as follows :
V=IR
V = 0.2 × 65
V = 13 volt
Answer:
1. 0.04 Ohms
2. 55 Ohms
3. 13 Volts
Explanation:
Penn Foster
Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J