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liberstina [14]
3 years ago
5

Velocity is a vector quantity that has both magnitude and direction using complete sentences describe the objects velocity comme

nt on both magnitude and the direction
Physics
1 answer:
sammy [17]3 years ago
8 0

An object travelling with a constant speed in different angles is termed as velocity of the object with speed being its magnitude and angles being its direction.

<u>Explanation:</u>

Velocity is a vector quantity which means it has a magnitude and direction. So magnitude is the quantitative amount of a given term and direction is its nature of flow.

If we consider a ball is thrown at a speed of 50 m/s in an angle of 30°, then the ball is exhibiting a velocity of 50 m/s with speed and direction as the angle of travelling direction in 30°. Thus, velocity consists of both magnitude as well as direction.

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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
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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}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

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

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

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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