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dimulka [17.4K]
3 years ago
9

In each cycle, a heat engine performs 710 J of work and exhausts 1480 J of heat. What is the thermal effiency?

Physics
1 answer:
Tomtit [17]3 years ago
6 0

Answer:

32%

Explanation:

For a heat engine, efficiency is work out divided by heat in:

η = Wₒ / Qᵢ

Since energy is balanced, heat in is the sum of work out and heat out:

Qᵢ = Wₒ + Qₒ

Therefore:

η = Wₒ / (Wₒ + Qₒ)

Given Wₒ = 710 J and Qₒ = 1480 J:

η = 710 / (710 + 1480)

η = 0.32

The thermal efficiency is 32%.

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An object is thrown with velocity v from the edge of a cliff above level ground. Neglect air resistance. In order for the object
musickatia [10]

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(C) greater than zero but less than 45° above the horizontal

Explanation:

The range of a projectile is given by R = v²sin2θ/g.

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What do the mitochondria do??
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A railroad car of mass 2.52 104 kg is moving with a speed of 3.86 m/s. It collides and couples with three other coupled railroad
aleksley [76]

Answer:

a)   v = 2.4125 m / s  , b)  Em_{f} / Em₀ = 0.89

Explanation:

a) This is an inelastic crash problem, the system is made up of the four carriages, so the forces during the crash are internal and the moment is conserved

Initial

          p₀ = m v₁ + 3 m v₂

Final

         p_{f} = (4 m) v

        p₀ =p_{f}

        m (v₁ + 3 v₂) = 4 m v

        v = (v₁ +3 v₂) / 4

Let's calculate

       v = (3.86 + 3 1.93) / 4

       v = 2.4125 m / s

b) the initial mechanical energy is

       Em₀ = K₁ + 3 K₂

       Em₀ = ½ m v₁² + ½ 3m v₂²

       

The final mechanical energy

         Em_{f} = K

         Em_{f} = ½ 4 m v²

The fraction of energy lost is

          Em_{f} / Em₀ = ½ 4m v² / ½ m (v₁² +3 v₂²)

          Em_{f} / Em₀ = 4 v₂ / (v₁² + 3 v₂²)

          Em_{f} / Em₀ = 4 2.4125² / (3.86² + 3 1.93²)

          Em_{f} / em₀ = 23.28 / 26.07

          Em_{f} / Em₀ = 0.89

6 0
3 years ago
The current in a circuit is tripled by connecting a 580 resistor in parallel with the resistance of the circuit. Determine the r
const2013 [10]

Answer:

1160 ohm

Explanation:

We are given that

R'=580 ohm

Current=3 I

We have to find the resistance of the circuit.

Let R be the resistance of circuit.

In parallel

\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}

Using the formula

\frac{1}{R_{eq}}=\frac{1}{580}+\frac{1}{R}=\frac{580+R}{580R}

R_{eq}=\frac{580R}{580+R}

In parallel combination,Potential difference across each resistance remains same.

V=IR

Using the formula

IR=3IR_{eq}

IR=3I\times \frac{580R}{580+R}

580+R=3\times 580=1740

R=1740-580=1160\Omega

6 0
3 years ago
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