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1 year ago

15

A certain heat engine extracts 1.30 kJ of heat from a hot temperature reservoir and discharges 0.70 kJ of heat to a cold tempera

ture reservoir. What is the efficiency of this engine?

2 answers:

skelet666 [1.2K]1 year ago

7 0

Answer:

46% (0.46)

Explanation:

temperature of hot reservoir (Th) = 1.3 kJ

temperature of COLD reservoir (Tc) = 0.7 kJ

Efficiency = 1 - (Tc/Th)

Efficiency = 1 - (0.7/1.3) = 0.46 = 0.46 x 100 = 46 %

Ede4ka [16]1 year ago

6 0

Answer:

46.2 %

Explanation:

heat in (Q_in) = 1.3kJ

heat out (Q_out) = 0.7kJ

work out = heat in - heat out = 1.3 - 0.7 = 0.6 kJ

efficiency = work out / heat in * 100%

= 0.6/1.3 * 100 % = 46.2 %

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Answer:

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At point 'P', there are no torques exerted by the F₁x and F₂x nor the weight of the bar as they all lie along the axis of rotation.

Therefore, the net torque by the forces $F_{1y}\ and\ F_{2y}$ will be zero. This gives,

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Therefore,

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We know,

$tan(\phi)=\frac{sin(\phi)}{cos(\phi)}\\\\cot(\phi)=\frac{cos(\phi)}{sin(\phi)}$

∴ $x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}$

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1 year ago

A varying force is given by F=Ae ^-kx, where x is the position;A and I are constants that have units of N and m^-1 , respectivel

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Answer:

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vredina [299]

Answer:

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