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lesya [120]
3 years ago
5

A real heat engine operates between temperatures Tc and Th. During a certain time, an amount Qc of heat is released to the cold

reservoir. During that time, what is the maximum amount of work Wmaxthat the engine might have performed?Express your answer in terms of Qc, Th, and Tc.
Physics
1 answer:
Ivanshal [37]3 years ago
4 0

Answer: Qc * (1 - Th/Tc)

Explanation:  

Maximum Work done = Qh * η

                                Wmax  = Qh * (1 - Tc/Th)

For reversible cycle of engine: Qc/Tc + Qh/Th = 0

                                                   Qh = -(Qc*Th)/Tc

Substituting Qh back into equation for work:

Wmax = -(Qc*Th)/Tc *  (1 - Tc/Th)

Wmax = -(Qc*Th)/Tc + Qc

Wmax = Qc * (1 - Th/Tc)

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(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit
Ainat [17]

Answer:

a

The number of fringe is  z  = 3 fringes

b

The  ratio is I = 0.2545I_o

Explanation:

a

 From the question we are told that

        The wavelength is  \lambda = 600 nm

        The distance between the slit is  d = 0.117mm = 0.117 *10^{-3} m

        The width of the slit is  a = 35.7 \mu m = 35.7 *10^{-6}m

let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

             z = \frac{d}{a}

Substituting values

             z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}  

             z  = 3 fringes

b

   From the question  we are told that the order  of the bright fringe is  n = 3

   Generally the intensity of  a pattern  is mathematically represented as

                 I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]

Where I_o is the intensity  of the  central fringe

 And  Generally  sin \theta = \frac{n \lambda }{d}

               I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]

               I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]

               I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]

                I = I_o (1)(0.2545)

                  I = 0.2545I_o

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Young's Modulus refers to changes in the a Volume b- Length c- Body layers
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Explanation:

Young' modulus is the ratio of normal stress to the longitudinal strain. Mathematically, it is given by :

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The mathematical definition of Young's modulus is given by :

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Where

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Answer:

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     Horiz component = 1704 cos 45 = 1204.9  m/s

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