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True [87]
3 years ago
10

Carnot engine A has an efficiency of 0.57, and Carnot engine B has an efficiency of 0.72. Both engines utilize the same hot rese

rvoir, which has a temperature of 615 K and delivers 1248 J of heat to each engine. Find (a), (b) the magnitude of the work produced by engine A and B respectively and (c), (d) the temperatures of the cold reservoirs that they use.
Physics
1 answer:
dem82 [27]3 years ago
4 0

Answer:

710.79 Joules

898.56 Joules

264.45 K

172.2 K

Explanation:

W = Work done

Q = Heat added = 1248 J

T_h = Temperature of hot reservoir = 615 K

T_c = Temperature of cold reservoir

Efficiencies of engine A and B

\eta_A = 0.57

\eta_B = 0.72

Efficiency is given by

\eta=\frac{W}{Q}\\\Rightarrow W=\eta Q\\\Rightarrow W=0.57\times 1248\\\Rightarrow W=711.36\ J

Work done by engine A is 710.79 Joules

\eta=\frac{W}{Q}\\\Rightarrow W=\eta Q\\\Rightarrow W=0.72\times 1248\\\Rightarrow W=898.56\ J

Work done by engine B is 898.56 Joules

Efficiency

\eta=1-\frac{T_c}{T_h}\\\Rightarrow 0.57=1-\frac{T_c}{615}\\\Rightarrow T_c=-0.43\times -615\\\Rightarrow T_c=264.45\ K

The temperature of the cold reservoir is 264.45 K

\eta=1-\frac{T_c}{T_h}\\\Rightarrow 0.72=1-\frac{T_c}{615}\\\Rightarrow T_c=-0.28\times -615\\\Rightarrow T_c=172.2\ K

The temperature of the cold reservoir is 172.2 K

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3 years ago
For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.
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Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

E = hv

E = \frac{hc}{\lambda}

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}

E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})

<u>E = 4.96 x 10³ eV</u>

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B)

The energy of a particle at rest is given as:

E = m_{0}c^2

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (9.1 x 10^{-31} kg)(3  x  10^8 m/s)^2\\

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C)

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E = m_{0}c^2

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E = (6.64 x 10^{-27} kg)(3  x  10^8 m/s)^2\\

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3 years ago
At the intersection of Texas Avenue and University Drive,
Zielflug [23.3K]

Answer:

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Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector \hat{i} pointing north and \hat{j} pointing east, the final velocity will be

\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )

\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

The final linear momentum will be:

\vec{P}_f = (m_{car}+ m_{truck}) * V_f

\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = (2.850 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )

As there are not external forces, the total linear momentum must be constant.

So:

\vec{P}_0= \vec{P}_f

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} ) 

so:

 \vec{P}_0= \vec{P}_f 

( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )  

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\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}  } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.

So, for the truck

m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = 21.93 \frac{m}{s}

And, for the car

950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}

v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}

v_{car}=19.524 \frac{m}{s}

5 0
3 years ago
A ruby laser delivers a 16.0-ns pulse of 4.20-MW average power. If the photons have a wavelength of 694.3 nm, how many are conta
stepan [7]

Answer:

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Explanation:

From the question we are told that

     The  amount of power delivered is  P  =  4.20 \  M W  =  4.20  *10^{6} \  W

      The  time taken is  t =  16.0ns  =  16.0 *10^{-9} \  s

       The  wavelength is  \lambda  =  694.3 \  nm =  694.3 *10^{-9} \  m

     

Generally the energy delivered is  mathematically represented as

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Where  h is the Planck's constant with value  h  =  6.262  *10^{-34} \  J \cdot  s

           c  is the speed of light with value  c =  3.0*10^{8} \  m/s

     

So  

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=>    n  =  2.347 *10^{17} \  photons

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ollegr [7]

My answer was incorrect, please disregard.

4 0
2 years ago
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