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3 years ago

12

Ann’s Retail, a women’s clothing store, hires female attendants to assist clients in the store’s dressing rooms. Larry, a male,

applies for, and is refused, a job as an attendant. Larry then sues Ann’s Retail for employment discrimination under Title VII. Against the suit, the store has

1 answer:

mojhsa [17]3 years ago

3 0

Answer:

A bona fide occupational qualification defense

Explanation:

Since the store is for women clothing, the retail may prefer to employ only female to assist the customers. Under a bona fide occupational qualification defense, an employer is allowed to discriminate if a characteristic is a necessity for the performance of the job and for the business. Therefore, the store has a bona fide occupational qualification defense.

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Find the phasor form of:

mixer [17]

Answer:

$I=24.598\angle 50.377$

Explanation:

A tension or current expressed in cosine form and with a positive sign can be converted directly into a phasor. This is done by indicating the tension and the offset angle:

$Acos(10\omega t +\phi)=A\angle \phi$

So:

$i(t)=10cos(10t+63)+15cos(10t-42)=10\angle 63 + 15\angle42=I$

You can sum the phasors simply using a calculator, however, let's do it manually:

Let's find the rectangular form of each phasor using the next formulas:

$A=\sqrt{a^2+b^2} \\\phi=arctan(\frac{b}{a})$

For $10\angle 63$

$63=arctan(\frac{b}{a} )\\\\tan(63)=\frac{b}{a} \\\\b=a*tan(63)$

$10=\sqrt{a^2+(a*tan(63))^2} \\\\10^2=a^2+a^2*(tan(63))^2\\\\Solving\hspace{3}for\hspace{3}a\\\\a=\sqrt{\frac{100}{1+tan(63)^2} } =4.539904997\\\\and\hspace{3}b\\b=\sqrt{100-a^2} =8.910065242$

So:

$Z_1=a+bj=4.539904997+8.910065242j$

For $15\angle 42$

$42=arctan(\frac{b_2}{a_2} )\\\\tan(42)=\frac{b_2}{a_2} \\\\b_2=a_2*tan(42)$

$15=\sqrt{a_2^2+(a_2*tan(42))^2} \\\\15^2=a_2^2+a_2^2*(tan(42))^2\\\\Solving\hspace{3}for\hspace{3}a_2\\\\a_2=\sqrt{\frac{225}{1+tan(42)^2} } =11.14717238\\\\and\hspace{3}b_2\\\\b_2=\sqrt{225-a^2} =10.0369591$

So:

$Z_2=a_2+b_2j=11.14717238+10.0369591j$

Hence:

$Z_T=Z_1+Z_2=(4.539904997+11.14717238)+(8.910065242+10.0369591)j\\Z_T=15.68707738+18.94702434j$

Finally:

$I=\sqrt{15.68707738^2+18.94702434^2} \angle arctan (\frac{18.94702434}{15.68707738} )=24.598\angle 50.377$

7 0

3 years ago

Mining is an example of this type of business

luda_lava [24]

Answer:

Mining would go under Industry organization.

5 0

4 years ago

Biomedical, electrical and civil engineering essay

ArbitrLikvidat [17]

Answer:

7th A good time for you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my

3 0

3 years ago

The equation for the velocity V in a pipe with diameter d and length L, under laminar condition is given by the equation V=Δpdsq

Citrus2011 [14]

Answer:

Given that

$V=\dfrac{\Delta Pd^2}{32\mu L}$

LHS of above given equation have dimension $[M^oL^{1}T^{-1}]$ .

Now find the dimension of RHS

Dimension of P = $[ML^{-1}T^{-2}]$ .

Dimension of d= $[M^{0}L^{1}T^{0}]$ .

Dimension of μ = $[ML^{-1}T^{-1}]$ .

Dimension of L= $[M^{0}L^{1}T^{0}]$ .

So

$\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}$

$\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]$

It means that both sides have same dimensions.

5 0

4 years ago

A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6

AnnZ [28]

Answer:

$\frac{du}{dt} = 1.515$

$\frac{dv}{dt} = 5.511$

Explanation:

The acceleration field is obtained by deriving the components in function of the time. That is to say:

$\frac{du}{dt}=2.05\cdot \frac{dx}{dt}+0.656\cdot \frac{dy}{dt}\\\frac{dv}{dt} = -2.18\cdot \frac{dx}{dt} -2.05\cdot \frac{dy}{dt}$

Where $\frac{dx}{dt} = u$ and $\frac{dy}{dt} = v$ .

The velocity components at given point are, respectively:

$u = 2.424$

$v=-5.266$

Lastly, the acceleration components are found:

$\frac{du}{dt} = 1.515$

$\frac{dv}{dt} = 5.511$

7 0

3 years ago