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Fofino [41]
3 years ago
12

A Carnot engine is operated between two heat reservoirs at temperatures of 550 K and 300 K . 1.) If the engine receives 6.70 kJ

of heat energy from the reservoir at 550 K in each cycle, how many joules per cycle does it reject to the reservoir at 300 K ? 2.)How much mechanical work is performed by the engine during each cycle?
Physics
1 answer:
Goshia [24]3 years ago
6 0

Answer:

a) 3654.5 joules per cycle is rejected to the reservoir at 300 K by the Carnot Engine.

b) Mechanical work performed by the engine during each cycle = 3045.5 J = 3.046 KJ

Explanation:

For a Carmot engine,

(Q꜀/T꜀) + (Qₕ/Tₕ) = 0

(Q꜀/T꜀) = - (Qₕ/Tₕ)

Q꜀ = - T꜀ × (Qₕ/Tₕ)

Qₕ = 6.70 KJ

Tₕ = 550 K

Q꜀ = ?

T꜀ = 300 K

Q꜀ = - 300 (6700/550) = - 3654.5 J = - 3.65 KJ

b) Mechanical work done by the Carnot engine is given by

W = |Qₕ| - |Q꜀| = 6700 - 3654.5 = 3045.5 J

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Answer:

c

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4 years ago
A curve of radius 55.1 m is banked so that a car of mass 1.6 Mg traveling with uniform speed 61 km/hr can round the curve withou
svp [43]

Answer:

θ = 28°

Explanation:

For this exercise We will use the second law and Newton, let's set a System of horizontal and vertical.

X axis

      Fₓ = m a

      Nₓ = m a

Where the acceleration is centripetal

      a = v² / r

The only force that we must decompose is normal, let's use trigonometry

      sin θ = Nₓ / N

      cos θ = N_{y} / N

      Nₓ = N sin θ

      N_{y} = N cos θ

Let's replace

     N sin θ = m v² / r

Y Axis

     N_{y} - W = 0

     N cos θ = mg

Let's divide the two equations of Newton's second law

     Sin θ / cos θ = v² / g r

     tan θ = v² / g r

     θ = tan⁻¹ (v² / g r)

We reduce the speed to the SI system

      v = 61 km / h (1000 m / 1 km) (1h / 3600 s) = 16.94 m / s

Let's calculate

     θ = tan⁻¹ (16.94 2 / (9.8 55.1)

     θ = tan⁻¹ (0.5317)

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3 0
3 years ago
Aaron walks 5 miles every day for exercise, leaving his front porch at 5:00 am. And returning to his porch at 6:00 am. What is t
taurus [48]

Answer:

Total displacement = 0

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He lives his front porch and still returns to his front porch.

Now, displacement is a vector quantity and as such, it is the distance between the initial point of movement and the final point of movement.

In this case the initial point is the same as the final point and thus the displacement is zero.

3 0
3 years ago
4. A vehicle
Gnom [1K]

Answer:

Average acceleration, a=2.77\ m/s^2

Explanation:

Given that,

Initial velocity, u = 0 km/h

Final velocity, v = 100 km/h = 27.77 m/s

Time, t = 10 s

We need to find the average acceleration of the vehicle. It is given by the change in velocity divided by time. Som,

a=\dfrac{v-u}{t}\\\\a=\dfrac{27.77-0}{10}\\\\a=2.77\ m/s^2

So, its average acceleration is 2.77\ m/s^2.

5 0
4 years ago
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