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Tcecarenko [31]

3 years ago

10

Weber theorized that we would eventually get stuck in ______, where we move from one formalized structure to the next, unable to

escape. a. A steel partition
b. A panopticon
c. An inhospitable planet
d. An iron cage

1 answer:

dsp733 years ago

8 0

Answer:

The correct answer is

d. An iron cage

Max Weber is well known for developing the theoretical concept of the iron cage

Explanation:

The theory of the iron was first presented by the sociologist Max Weber in the work The Protestant Ethic and the Spirit of Capitalism.

Here Weber clarified that with the diminishing effect of Protestantism within our social life the capitalist system remained as well as the bureaucratic principles and social structures that came about along with came along with it, such the worldviews and values becomes major controlling factors in socil life which is tantamount to living in a strong house maade of steel

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If the third charge (–| q 3 |) is placed at point P, but not held fixed, it will experience a force and accelerate away from the

Mazyrski [523]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

The potential of this system is $U=6.75*10^{-7}J$

B

The electric potential at point p is $V_p= -900V$

C

The work required is $W= 9*10^{-7}J$

D

The speed of the charge is $v=600m/s$

Explanation:

A sketch to explain the question is shown on the second uploaded image

Generally the potential energy for a system of two charges is mathematically represented as

$U = \frac{kq_1 q_2}{d}$

where k is the electrostatic constant with a value of $k = 9*10^9 N m^2 /C^2$

q is the charge with a value of $q = 1*10^{-9}C$

d is the distance given as $d =5m$

Now we are given that $q_1 = q$ and $q_2 = 3q$ and

Now substituting values

$U = \frac{9*10^9 *1*10^{-9} * 3*10^{-9}}{5}$

$U=6.75*10^{-7}J$

The electric potential at point P is mathematically obtained with the formula

$V_p = V_{-q} + V_{-3q}$

I.e the potential at $q_1$ plus the potential at $q_2$

Now potential at $q_1$ is mathematically represented as

$V_{-q} = \frac{-kq}{s}$

and the potential at $q_2$ is mathematically represented as

$V_{-3q} = \frac{-3kq}{s}$

Now substituting into formula for potential at P

$V_p = \frac{-kq}{s} + \frac{-3kq}{s} = -\frac{4kq}{s}$

$= \frac{4*9*10^9 *1*10^{-9}}{4*10^{-2}}$

$V_p= -900V$

The Workdone to bring the third negative charge is mathematically evaluated as

$W =\Delta U = \frac{kq_1q_3}{s} + \frac{kq_2q_3}{s}$

$= \frac{kq*q}{s} + \frac{kq*3q}{s}$

$= \frac{4kq^2}{s}$

$= \frac{4* 9*10^9 * (1*10^{-9})^2}{4*10^{-2}}$

$W= 9*10^{-7}J$

From the Question are told that the charge $q_3$ would a force and an acceleration which implies that all its potential energy would be converted to kinetic energy.This can be mathematically represented as

$\Delta U = W = \frac{1}{2} m_{q_3} v^2$

$9*10^{-7} = \frac{1}{2} m_{q_3} v^2$

Where $m_{q_3} = 5.0*10^{-12}kg$

Now making v the subject we have

$v = \sqrt{\frac{9*10^{-12}}{5*10^{-12}*0.5} }$

$v=600m/s$

8 0

3 years ago

Ok plz answer and tell me how to do it

kirza4 [7]

Answer: 25N

method: total force in the right hand direction is 100N and total force in the left hand direction is 125N. To get the net force, we add forces if they are in the same direction and substract if they are in opposite directions. since 100N and 125N are in opposite directions, we substract the larger value from the smaller value. Then we get 25N in the left hand direction as the final answer.

4 0

3 years ago

2. Adelia holds a shiny steel spoon with its back (convex surface) facing her eyes at a distance

My name is Ann [436]

Answer:

(a) The convex mirror image, is always upright at all positions, while images formed by concave mirrors are always inverted when the object distance from the mirror is more than the mirrors focal length.

(b) An upright image is not seen for object at a distance from a concave mirror further than the focal length of the mirror, which is the spoon in the question

Therefore, the location of her eyes of approximately, 30 cm, from the mirror is more than the mirror's focal length

Explanation:

3 0

3 years ago

A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that

lina2011 [118]

Answer:

0.006075Joules

Explanation:

The final kinetic energy of the system is expressed as;

KE = 1/2(m1+m2)v²

m1 and m2 are the masses of the two bodies

v is the final velocity of the bodies after collision

get the final velocity using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

0.12(0.45) + 0/12(0) = (0.12+0.12)v

0.054 = 0.24v

v = 0.054/0.24

v = 0.225m/s

Get the final kinetic energy;

KE = 1/2(m1+m2)v

KE = 1/2(0.12+0.12)(0.225)²

KE = 1/2(0.24)(0.050625)

KE = 0.12*0.050625

KE = 0.006075Joules

Hence the final kinetic energy of the system is 0.006075Joules

5 0

3 years ago

A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student use

polet [3.4K]

The net force of the object is equal to the force applied minus the force of friction.
Fnet = ma = F - Ff
12 kg x 0.2 m/s² = 15 N - Ff
The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
Ff = 12.6 N = k(12 kg)(9.81 m/s²)
The value of k is equal to 0.107.

7 0

3 years ago