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nadya68 [22]
2 years ago
13

A heat engine extracts 42. 53 kj from the hot reservior and exhausts 17. 69 kj into the cold reservior. what is the work done?

Physics
1 answer:
ludmilkaskok [199]2 years ago
6 0

The answer is  24.84kJ.

We apply the expression for the work done by the heat engine is,

W=E_{i n}-E_{o u t}. Putting all given values in the equation we get the final answer.

What is heat engine?

  • A heat engine is a machine that uses heat to generate power. It draws heat from a reservoir, uses that heat to produce work, such as move a piston or lift weights, and then releases that heat energy into the sink.
  • We are given:The heat input is $E_{i n}=42.53 \mathrm{~kJ}$. The heat output is $E_{o u t}=17.69 \mathrm{~kJ}$.
  • The expression for the work done by the heat engine is,W=E_{i n}-E_{o u t}
  • Substituting the given values in the above expression, we will getW =42.53 \mathrm{~kJ}-17.69 \mathrm{~kJ}=24.84kJ.
  • Thus, the work done by the heat engine is 24.84kJ.

To learn more about heat engine visit: brainly.com/question/15735984

#SPJ4

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Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is  v_A= 4m/s

Explanation:

From the question we are told that

    The length of the string is  L = 1m

     The initial speed of block A is u_A

     The final speed of block A is  v_A = \frac{1}{2}u_A

      The initial speed of block B is u_B = 0

      The mass of block A  is  m_A = 7kg  gh

      The mass of block B is  m_B  = 2 kg

According to the principle of conservation of momentum

       m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}

Since block B at initial is at rest

       m_A u_A  = m_Bv_B + m_A \frac{u_A}{2}

      m_A u_A  - m_A \frac{u_A}{2} = m_Bv_B

          m_A \frac{u_A}{2} = m_Bv_B

  making v_B the subject of the formula

             v_B =m_A \frac{u_A}{2 m_B}

Substituting values

               v_B =\frac{7 u_A}{4}  

This v__B is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity  of v__B' at  the top of the vertical circle  

 The angular centripetal acceleration  would be mathematically represented

                   a= \frac{v^2_{B}'}{L}

Note that  this acceleration would be toward the center of the circle

      Now the forces acting at the top of the circle can be represented mathematically as

         T + mg = m \frac{v^2_{B}'}{L}

    Where T is the tension on the string

  According to the law of energy conservation

The energy at  bottom of the vertical circle   =  The energy at the top of

                                                                                the vertical circle

   This can be mathematically represented as

                 \frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L

From above  

                (T + mg) L = m v^2_{B}'

Substitute this into above equation

             \frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L  + mg 2L  

             \frac{49 mv_A^2}{16}  = \frac{1}{2} (T + mg) L + mg 2L

          \frac{49 mv_A^2}{16}  = T + 5mgL

The  value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is  zero

        This is mathematically represented as

                      \frac{49 mv_A^2}{16}  = 5mgL

making  v_A the subject

            v_A = \sqrt{\frac{80mgL}{49m} }

substituting values

          v_A = \sqrt{\frac{80* 9.8 *1}{49} }

              v_A= 4m/s

     

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