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3 years ago

Which friction-reducing technologies are used in the Variable Compression Turbo Engine?

1 answer:

4 0

Friction-reducing technologies used in the Variable Compression Turbo Engine are Diamond-like coating on valve lifters, micro finishing on crankshaft and camshaft and mirror bore coating on cylinder wall.

<u>Explanation:</u>

Variable compression is a technology to adjust the compression of an internal combustion engine while the engine is in operation. At this time friction may occur that need to be reduced. To reduce this friction some technologies are used like

Diamond-like coating on valve lifters
Micro finishing on crankshaft and camshaft
Mirror bore coating on cylinder wall

A hydrogen free diamond like carbon coating is applied to an engine valve lifter to reduce mechanical loss. Micro finishing on crankshaft and camshaft achieves improvement in geometric parameters such as roundness. Mirror bore coating on cylinder wall raises energy efficiency by reducing the friction inside the engine.

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Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

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4 years ago

A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of t

grandymaker [24]

Answer:

13.23J

Explanation:

PE = m*g*h

PE = (3 kg ) * (9.8 m/s/s) * (0.45 m)

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3 years ago

Read 2 more answers

Select all that apply. Which of the following would increase the conductivity of a connecting wire?

lord [1]

What are your choices

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4 years ago

Which statement is true of the energies of the bonds that break and form during the reaction? which statement is true of the ene

Inga [223]

<span>The statement that is true of the energies required to break bonds is this:

the energy needed to break the required bonds is greater than the energy released when the new bonds form.

Bonds that have been formed during specific reactions require much energy to be broken down again because of the closed atomic bond that have formed during the process.</span>

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3 years ago

Read 2 more answers

Juliette sets the initial velocity to +10 m/s, the acceleration to zero, and clicks “start.” How can Shakina describe the subseq

ICE Princess25 [194]

Explanation :

It is given that :

The initial velocity of the car is 10 m/s.

Juliette sets the acceleration to zero.

We know that, acceleration $a=\frac{v-u}{t}$ \

where,

u is the initial velocity and v is the final velocity.

So, final velocity will become 10 m/s.

Hence, the car will move with the constant velocity i.e. 10 m/s.

So, Shakina described that the car will move with the constant velocity of 10 m/s as acceleration is zero.

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3 years ago