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galina1969 [7]
3 years ago
12

The diagram shows the four primary steps in the production of work in a four-stroke heat engine.

Physics
1 answer:
Alika [10]3 years ago
6 0

Answer:

D- Z, X, Y, W

Explanation:

Correct on edge

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A ferry coming into port is sailing at 12 m/s. It takes 2.5km to come to rest in the port. Calculate the deceleration of the fer
pogonyaev

Answer:    -0.0288 m/s^2

Explanation:

Let's suppose that the ferry decelerates at a constant rate A (deceleration is an acceleration in the opposite direction to the original motion of an object)

Then the acceleration equation of the ferry will be:

a(t) = -A

(the negative sign is because this acceleration is in the opposite direction with respect to the movement of the ferry)

To get the velocity equation of the ferry, we need to integrate with respect to the time, t, we will get:

v(t) = -A*t + v0

where v0 is the initial velocity of the ferry, v0 = 12m/s.

v(t) = -A*t + 12m/s

For the position equation of the ferry we need to integrate again over time:

p(t) = (-A/2)*t^2 + (12m/s)*t + p0

Where p0 is the initial position of the ferry, in this case, it can be zero, because it will depend on where we put the origin on our coordinate axis.

then p0 = 0m

P(t) = (-A/2)*t^2 + (12m/s)*t

The ferry will come to rest at the moment when it's velocity is equal to zero, this will happen when:

v(t) = 0m/s = -A*t + 12m/s

We need to find the value of t.

A*t = 12m/s

t = (12m/s)/A

Now we can replace this in the position equation because we know that the ferry needs 2.5 km or 2500 meters to come to rest.

p(  (12m/s)/A) = 2500m =  (-A/2)*( (12m/s)/A)^2 + (12m/s)*((12m/s)/A)

2500m = (-72 m^2/s^2)/A + (144m^2/s^2)/A

2500m = (72 m^2/s^2)/A

2500m*A = (72 m^2/s^2)

A = (72 m^2/s^2)/2500m = 0.0288 m/s^2

and the acceleration of the ferry was -A, then the acceleration of the ferry is:

-0.0288 m/s^2

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2 years ago
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

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On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

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The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

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Explanation: In places where a plate faced resistance to its movement, it would fold upward and create mountains. Hope this helped! :)

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