The Otto-cycle engine in a Mercedes-Benz SLK230 has a compression ratio of 8.8.Part A. What is the ideal efficiency of the engine? Use γ=1.40.
Part B. The engine in a Dodge Viper GT2 has a slightly higher compression ratio of 9.6. How much increase in the ideal efficiency results from this increase in the compression ratio?

Answers

Answer 1
Answer:

Answer:

58.1%

59.533%

1.433%

Explanation:

(V_1)/(V_2) = Compression ratio

\gamma = Specific heat ratio = 1.4

Efficiency in the Otto cycle is given by

\eta=1-(1)/(\left((V_1)/(V_2)\right)^(\gamma-1))\n\Rightarrow \eta=1-(1)/(\left(8.8\right)^(1.4-1))\n\Rightarrow \eta=0.581

The ideal efficiency of the engine is 58.1%

\eta=1-(1)/(\left((V_1)/(V_2)\right)^(\gamma-1))\n\Rightarrow \eta=1-(1)/(\left(9.6\right)^(1.4-1))\n\Rightarrow \eta=0.59533

The ideal efficiency of the engine is 59.533%

Increase in efficiency is 59.533-58.1 = 1.433%


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C. an X-ray.


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Which terrestrial planet experiences the shortest year?A. Mars
B. Earth
C. Venus
D. Mercury

Answers


Being the planet closest to the sun, Mercury must be the one
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How much current will a 240 V heater with 20 ohms of resistance draw?

Answers

Answer:

How much current will a 240 V heater with 20 ohms of resistance draw?=12A

two small balls are suspended on parallel threads of the same length so that they touch each other in the vertical position. the mass of the first ball is 0.4kg and the mass of the second ball is 200g. the first ball is deflected so that its centre of mass rises to a height of 6cm and it is then released. to what maximal height will the 200g ball rise if the collision is elastic?

Answers

Answer: 12cm

Explanation:

Since first ball is raised to a height h1= 6cm, it gains a potential energy(PE1). On release, this PE1 is converted to kinetic Energy (KE1). This KE1 impact it energy on ball 2 ans it rise also. Since collision is elastic.

KE1 = KE2 ....> PE1 = PE2

m1 ×g × h1 = m2 × g× h2

h2 = m1×h1/m2

h2 = 0.4 × 0.06/0.2

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If you were standing at the center of curvature in front of a concave mirror, what image would be projected?A. The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

B.The image would be upright, would look shorter than you, and would be closer to the mirror than you are.

C.The image would be upside down, would look shorter than you, and would be closer to the mirror than you are.

D.The image would be upright, would look as tall you, and would be at the same distance from the mirror as you are.

Answers

The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

Answer:

A. The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

Explanation:

As we know by the mirror formula

(1)/(d_i) + (1)/(d_o) = (1)/(f)

now we know that object is placed at distance equal to the radius of mirror

d_o = -R

also we know that focal length of mirror is half of the radius of mirror

f = (R)/(2)

now we have

(1)/(d_i) - (1)/(R) = -(2)/(R)

so we have

d_i = -R

so image will be at same position as that the position of object and it is inverted in position so correct answer will be

A. The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

A high pressure center is generally characterized by

Answers

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