Why do societies stratify?

Answers

Answer 1

Answer:

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.

Answer 2

Answer:

Final answer:

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.

Explanation:

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.

Learn more about Social Stratification here:

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You have a grindstone (a disk) that is 105.00 kg, has a 0.297-m radius, and is turning at 71.150 rpm, and you press a steel axe against it with a radial force of 46.650 N. Assuming the kinetic coefficient of friction between steel and stone is 0.451. How many turns will the stone make before coming to rest?

Answers

Answer:

3.27 turns

Explanation:

To find how many turns (θ) will the stone make before coming to rest we will use the following equation:

$\omega_(f)^(2) = \omega_(0)^(2) + 2\alpha*\theta$

Where:

$\omega_(f)$ : is the final angular velocity = 0

$\omega_(0)$ : is the initial angular velocity = 71.150 rpm

α: is the angular acceleration

First, we need to calculate the angular acceleration (α). To do that, we can use the following equation:

$\alpha = (\tau)/(I)$

Where:

I: is the moment of inertia for the disk

τ: is the torque

The moment of inertia is:

$I = (mr^(2))/(2)$

Where:

m: is the mass of the disk = 105.00 kg

r: is the radius of the disk = 0.297 m

$I = (105.00 kg*(0.297 m)^(2))/(2) = 4.63 kg*m^(2)$

Now, the torque is equal to:

$\tau = -F x r = -\mu*F*r$

Where:

F: is the applied force = 46.650 N

μ: is the kinetic coefficient of friction = 0.451

$\tau = -\mu*F*r = -0.451*46.650 N*0.297 m = -6.25 N*m$

The minus sign is because the friction force is acting opposite to motion of grindstone.

Having the moment of inertia and the torque, we can find the angular acceleration:

$\alpha = (-6.25 N*m)/(4.63 kg*m^(2)) = -1.35 rad/s^(2)$

Finally, we can find the number of turns that the stone will make before coming to rest:

$0 = \omega_(0)^(2) + 2\alpha*\theta$

$\theta = -((\omega_(0))^(2))/(2\alpha) = -((71.150 (rev)/(min))^(2))/(2*(-1.35 (rad)/(s^(2)))*(1 rev)/(2\pi rad)*((60 s)^(2))/((1 min)^(2))) = 3.27 rev = 3.27 turns$

I hope it helps you!

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with another whose net charge is + 4Q. We move the +Q and +4Q charges to be 3 times as far apart as they were. What is the magnitude of the force on the +4Q charge ?A. F

B. 4F

C. 4F/3

D. 4F/9

E. F/3

Answers

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

$F=(kQ^2)/(r^2)$ ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

$F'=(kQ* 4Q)/((3r)^2)\n\nF'=(4kQ^2)/(9r^2)$ .....(2)

Dividing equation (1) and (2), we get :

$(F)/(F')=((kQ^2)/(r^2))/((4kQ^2)/(9r^2))\n\n(F)/(F')=(kQ^2)/(r^2)* (9r^2)/(4kQ^2)\n\n(F)/(F')=(9)/(4)\n\nF'=(4F)/(9)$

Hence, the correct option is (d) i.e. " 4F/9"

Final answer:

The magnitude of the force on the +4Q charge, after replacing one of the original +Q charges and moving the charges three times farther apart, is calculated to be 4F/9 using Coulomb's Law. Therefore, the correct answer is D.

Explanation:

The magnitude of the electrostatic force between two charges can be described by Coulomb's Law, which states that F = k × (q1 × q2) / r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the centers of the two charges. Originally, two objects each with charge +Q exert a force of magnitude F on each other. After one charge is replaced with a +4Q charge and they are moved to be three times as far apart, the force on the +4Q charge can be calculated using the modified version of Coulomb's Law that takes into account the new charges and distance.

Using the original scenario as a reference, where F = k × (Q × Q) / r^2, when the charge is replaced and the distance is tripled, the new force F' = k × (Q × 4Q) / (3r)^2 = 4kQ^2 / 9r^2. By comparing F' with F, we find that F' = (4/9)F. Thus, the magnitude of the force on the +4Q charge is 4F/9.

Select all the statements regarding electric field line drawings that are correct. Group of answer choices:
1. Electric field lines are the same thing as electric field vectors.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space.
3. The number of electric field lines that start or end at a charged particle is proportional to the amount of charge on the particle.
4. The electric field is strongest where the electric field lines are close together.

Answers

Answer:

All statement are correct.

Explanation:

1. Electric field lines are the same thing as electric field vectors, electric field are mathematically vectors quantity. These vectors point in the direction in which a positive test charge would move.

2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space. Yes it is correct tangent drawn at any point on these lines gives the direction of electric filed at that point.

3. The number of electric field lines that start or end at a charged particle is proportional to the magnitude of charge on the particle, is a correct statement.

4.The electric field is strongest where the electric field lines are close together, again a correct statement as relative closeness of field lines indicate a stronger strength of electric field.

Hence we can say that all the statement are correct.

Halogen lightbulbs allow their filaments to operate at a higher temperature than the filaments in standard incandescent bulbs. For comparison, the filament in a standard lightbulb operates at about 2900K, whereas the filament in a halogen bulb may operate at 3400K. Which bulb has the higher peak frequency? Calculate the ratio of the peak frequencies. The human eye is most sensitive to a frequency around 5.5x10^14 Hz. Which bulb produces a peak frequency close to this value?

Answers

Answer:

Halogen

0.85294

Explanation:

c = Speed of light = $3* 10^8\ m/s$

b = Wien's displacement constant = $2.897* 10^(-3)\ mK$

T = Temperature

From Wien's law we have

$\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(2900)\n\Rightarrow \lambda_m=9.98966* 10^(-7)\ m$

Frequency is given by

$\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(9.98966* 10^(-7))\n\Rightarrow \nu=3.00311* 10^(14)\ Hz$

For Halogen

$\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(3400)\n\Rightarrow \lambda_m=8.52059* 10^(-7)\ m$

Frequency is given by

$\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(8.52059* 10^(-7))\n\Rightarrow \nu=3.52088* 10^(14)\ Hz$

The maximum frequency is produced by Halogen bulbs which is closest to the value of $5.5* 10^(14)\ Hz$

Ratio

$(3.00311* 10^(14))/(3.52088* 10^(14))=0.85294$

The ratio of Incandescent to halogen peak frequency is 0.85294

The spring is used in a spring gun to project a 10.0 g plastic ball. In the gun, the spring is compressed by 10.0 cm before the gun is fired. How fast is the ball traveling after the gun is fired?

Answers

Answer:

5.831 m/s

Explanation:

According to the work-energy law,

Work done between two points = Change in kinetic energy between the two points.

Since the plastic ball is initially at rest, its initial kinetic energy is 0 since the initial velocity = 0

Work done by the spring = ∫ F.dx

The spring is compressed by 10 cm, so, we integrate from -0.1 m to 0 m

Fₓ(x) = (-30.0 N/m)x+ (60.0 N/m²)x²

F = -30x + 60x²

W = ∫ F.dx = ∫ (-30x + 60x²) dx

W = [- 15x² + 20x³]⁰₋₀.₁ = 0 - [- 15(0.01) + 20(-0.001)] = 0.17 J

W = ΔKE

ΔKE = (mv²/2) - 0

mv²/2 = 0.17

m = 10 g = 0.01 kg

0.01 v² = 0.34

v² = 34

v = 5.831 m/s

A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.46 rad/s2. Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of -31.4 rad/s. While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

Answers

Answer:

The time for the change in the angular velocity to occur is 14.08 secs

Explanation:

From the question,

the angular acceleration is - 4.46 rad/s²

Angular acceleration is given by the formula below

$\alpha =(\omega -\omega _(o) )/(t - t_(o) )$

Where $\alpha$ is the angular acceleration

$\omega$ is the final angular velocity

$\omega _(o)$ is the initial angular velocity

$t$ is the final time

$t_(o)$ is the initial time

From the question

$\alpha$ = - 4.46 rad/s²

$\omega _(o)$ = 0 rad/s (starting from rest)

$\omega$ = -31.4 rad/s

$t_(o)$ = 0 s

Now, we will determine t

From $\alpha =(\omega -\omega _(o) )/(t - t_(o) )$ , then

$-4.46 = (-31.4 - 0)/(t - 0)$

$-4.46 = (-31.4)/(t)$

$t = (-31.4)/(-4.46)$

t = 7.04 secs

This is the time spent in one direction,

Since the angular displacement of the wheel is zero ( it returned to its initial position), then the time required for the change in the angular velocity will be twice this time, that is 2t

Hence,

The time is 2×7.04 secs = 14.08 secs

This is the time for the change in the angular velocity to occur.

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