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UNO [17]
3 years ago
9

Jet fighter planes are launched from aircraft carriers with the aid of their own engines and a catapult. If in the process of be

ing launched from rest, the thrust of a jet's engines is 2.50 x 10^5 N and after moving through a distance of 90 m the plane lifts off with a kinetic energy of 5.40 x 10^7 J, what is the work done on the jet by the catapult?
Physics
1 answer:
ANTONII [103]3 years ago
5 0

Answer:

W=315 x 10⁵ J

Explanation:

Given that

F= 2.5 x 10⁵ N

d= 90 m

K.E.=5.4 x 10⁷ J

We know that work done by all force is equal to the change in kinetic energy

Lets take work done by catapult is W

W + F.d= K.E.

W= 5.4 x 10⁷ -  2.5 x 10⁵  x 90 J

W= (540 - 25 x 9) 10⁵ J

W=315 x 10⁵ J

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</span> a=v^2/r 
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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
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}}

\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

4 0
3 years ago
Work occurs when
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The answer is the FIRST OPTION 
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An ideal refrigerator does 130. 0 j of work to remove 780. 0 j of heat from its cold compartment during each cycle. what is the
zheka24 [161]

The refrigerator's coefficient of performance is 6.

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As the inside temperature of the refrigerator decreases, its coefficient of performance decreases. The coefficient of performance (COP) of refrigeration is always more than 1.

The heat produced in the cold compartment, H = 780.0 J

Work done in ideal refrigerator, W = 130.0 J

Refrigerator's coefficient of performance = H/W

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Therefore, the refrigerator's coefficient of performance is 6.

Energy conservation requires the exhaust heat to be = 780 + 130

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