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Pepsi [2]
3 years ago
10

How do the early efforts of women during the suffrage movement compare to the later years

Physics
1 answer:
Nat2105 [25]3 years ago
7 0

The idea that women's liberation is also men's liberation arouses the awakening of feminism. In 1966 Frieden and other feminists founded the National Organization for Women. Then other women's organizations for equal rights were formed in the United States and Western Europe. These organizations sought to overturn discriminatory laws and practices that forced women to hold down their status, such as discrimination in matters of contract and property ownership, employment and wages, the treatment of wage income, and issues relating to sex and child-bearing. The growing feminist movement more broadly seeks to change social stereotypes that women are more vulnerable, passive and dependent, less rational and emotionally reactive than men. Feminism strives for greater freedom for women in the workplace and makes them financially and psychologically independent of men, if they wish. Feminists criticized society's general emphasis on women as objects of sexual desire and sought to raise women's awareness and expand their opportunities in order to achieve equality with men. Another objective of feminism is to promote women's participation in all areas of political decision-making and social life.

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A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to
Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0
3 years ago
The barometric pressure in breckenridge, colorado (elevation 9600 feet) is 580 mm hg. how many atmospheres is this?
Maksim231197 [3]

1 atmospheric pressure = 760.0 mm Hg

Thus 580 mm Hg = (580 mm Hg/(760 mm Hg/atm))

= 0.763 atm


7 0
3 years ago
What is the pressure 100m below the surface of the sea , if the density of sea water is 1150kg/m3
Olin [163]
The pressure at 100 meters below the surface of sea water with a density of 1150kg is 145.96 psi.
4 0
3 years ago
Consider a block of mass m attached to two springs, one on the left with spring constant k1 and one on the right with spring con
Andrew [12]

Answer:

T= 2\pi\times sqrt(m/(k1+k2))

Explanation:

When the block is displaced by x units

F= spring force

two springs are connected parallel

F =-k_1x - k_2x

Writing Newtons second law, F = ma

-k_1x - k_2x =ma

-k_1x - k_2x = mx''

a= x" ( differentiating x w.r.t time twice)

x''+(k_1/m + k_2/m) x=0

this the standard form of equation of oscillation spring mass system

This is the differential equation, x'' means that double differentiation of x , i.e, x'' is acceleration

since,  Period T=2\pi\sqrt{\frac{m}{K_{eq.}} }

therefore,

T= 2\pi\times sqrt(m/(k1+k2))

3 0
3 years ago
You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.
Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x  Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

F_{a} =\frac{F_{f}+F_{i}  }{2}  Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx :  displacement

W = F_{a} (x_{f} -x_{i} )   :Formula (3)

x_{f} :  final deformation

x_{i}  :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N

F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N

We calculate average force applying formula (2):

F_{a} =\frac{15.4N+6.2N}{2} = 11 N

We calculate the work done on the spring  applying formula (3) :         :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

E_{pf} =\frac{1}{2} *k*x_{f}^{2}

E_{pi} =\frac{1}{2} *k*x_{i}^{2}

E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J

E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J

W=ΔEp=  5.39 J-0.99 J = 4.4J

:

4 0
3 years ago
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