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ziro4ka [17]
3 years ago
11

stimate the maximum efficiency of an automobile engine that has a compression ratio of 5:1.0. Assume the engine operates accordi

ng to the Otto cycle and assume γ = 1.4.
Engineering
1 answer:
Fed [463]3 years ago
5 0

Answer:

Efficiency based on Otto cycle.

Effotto = 47.47%

Explanation:

Efficiency based on Otto cycle.

effotto = 1 – (V2 / V1)^γ-1

effotto = 1 – (1 / 5)^1.4 - 1

effotto = 47.47%

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A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po
Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: 4.3\cdot 10^{-15} J

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

V(r)=\frac{kq}{r}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

q_1=+2.00\mu C=+2.00\cdot 10^{-6}C

and is located at the origin (x=0, y=0)

Charge 2 is

q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

r_{1A}=0.40 m

So the potential due to charge 2 is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V

The distance of point A from charge 2 is

r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m

So the potential due to charge 1 is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V

Therefore, the net potential at point A is

V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m

So the potential due to charge 1 at point B is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V

The distance of charge 2 from point B is

r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m

So the potential due to charge 2 at point B is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V

Therefore, the net potential at point B is

V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V

So the potential difference is

V_B-V_A=-29,500 V-(-2700 V)=-26,800 V

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

W=q\Delta V

where

q is the charge of the particle

\Delta V is the potential difference

In this problem, we have:

q=-1.6\cdot 10^{-19}C is the charge of the electron

\Delta V=-26,800 V is the potential difference

Therefore, the work required on the electron is

W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J

4 0
3 years ago
A cruise missile under test is moving horizontally at Ma =2 in the atmosphere at an elevation of 2000 m (Air temperature is 2 °C
telo118 [61]

Answer: the half-angle "alpha" of the Mach cone = 30⁰

Explanation:

To calculate the  half-angle "alpha" of the Mach cone.

we say ;

Sin∝ = 1 / Ma

given that Ma = 2

now we substitute

Sin∝ = 1 / 2

Sin∝ = 0.5

∝ = Sin⁻¹ 0.5

∝ = 30⁰

Therefore, the half-angle "alpha" of the Mach cone is 30⁰

3 0
3 years ago
I have a Dutch oven that looks like this what do I do?
dangina [55]

Answer:

baking soda and vinegar dish soap

Explanation:

it will create a bubbles and let it sit for 3 hours and it will go away

3 0
3 years ago
Bind hole, 38 diameter, .50 deep
agasfer [191]

Answer:

59.69021

Explanation:

38/.5 x 3.14159

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Consider the 'scope trace showing the input signal to an rc filter (the one having the larger amplitude) and the output of the f
yulyashka [42]

The corner frequency of this filter in hz is Directly switched the documents to one-of-a-kind Cloud Multi-Regional Storage bucket places in US, EU, and Asia the usage of APIs over HTTP(S).

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It is geo-redundant, this means that Cloud Storage shops your information redundantly in as a minimum geographic location separated via way of means of as a minimum one hundred miles in the multi-local vicinity of the bucket.

Multi-Regional Storage is suitable for storing information this is often accessed ("hot" objects), along with serving internet site content, interactive workloads, or information helping cell and gaming applications. Multi-Regional Storage information has the maximum availability as compared to different garage classes.

Read more about the amplitude:

brainly.com/question/413740

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6 0
2 years ago
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