1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
AleksAgata [21]
3 years ago
9

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

Physics
1 answer:
Grace [21]3 years ago
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.

You might be interested in
How much time would it take for the sound of thunder to travel 1,500 meters if sound travels at a speed of 330 m/sec?
Fittoniya [83]

(1,500 meters) x (1 sec / 330 meters) = (4 and 18/33) seconds

                                                                 (4.55 sec, rounded)

8 0
3 years ago
Read 2 more answers
A car travels 20 km South, then turns and travels 30 km East? What is the total displacement?
Ne4ueva [31]

Answer:

Explanation:36.05 km

Given

First car travels r_1=20\ km South

then turns and travels r_2=30\ km east

Suppose south as negative y axis and east as positive x axis

So, r_1=-20\hat{j}

r_2=30\hat{i}

Displacement is the shortest between initial and final point

Dispalcement=r=r_1+r_2

Displacement=-20\hat{j}+30\hat{i}

Displacement=30\hat{i}-20\hat{j}

Magnitude =\sqrt{30^2+(-20)^2}

Magnitude=36.05\ km

4 0
3 years ago
What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential di
yan [13]
What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10^4v ?
3 0
1 year ago
A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75
8_murik_8 [283]

Complete Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer:

\theta=274rev

Explanation:

From the question we are told that:

Angular velocity \omega=510rpm

Mass m=40.kg

Diameter d 75=>0.75m

Off Time t=40.0s

Oscillation at Power off N=210

Generally the equation for Angular displacement is mathematically given by

 \theta_{\infty}=\frac{w+w_0}{t}t

 w=\frac{2*\theta_{\infty}}{t}-w_0

 w=\frac{28210}{40*(\frac{1}{60})}-510

 w=120rpm

Generally the equation for Time to come to rest is mathematically given by

 t=(\frac{\omega_0}{\omega_0-\omega})t

 t=(\frac{510}{510-120rpm})(40.0)(\frac{1}{60})

 t=0.87min

Therefore Angular displacement is

 \theta =(\frac{120+510}{2})0.87

 \theta=274rev

6 0
2 years ago
Two charges are located in the xx–yy plane. If q1=−4.10 nCq1=−4.10 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.080 m)
Gala2k [10]

Answer:

Explanation:

Due to first charge , electric field at origin will be oriented towards - ve of y axis.

magnitude

Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j

= - 31.6 j N/C

Due to second charge electric field at origin

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.2²+ .6²

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.8

= 18 N/C

It is making angle θ where

Tanθ = .6 / 1.2

= 26.55°

this field in vector form

= - 18 cos 26.55 i - 18 sin26.55 j

= - 16.10 i - 8.04 j

Total field

= - 16.10 i - 8.04 j + ( - 31.6 j )

= -16.1 i - 39.64 j .

Ex = - 16.1 i

Ey = - 39.64 j .

8 0
3 years ago
Other questions:
  • How do you know when a penalty has been called?
    15·1 answer
  • A noisy toy is placed in a glass jar. All the air is then slowly removed from the jar. How does this affect the sound you hear?
    7·1 answer
  • When two resistors are connected in parallel across a battery of unknown voltage, one resistor carries a current of 3.3 a while
    10·1 answer
  • Which best explains the relationship between heat energy and temperature?
    10·1 answer
  • Which substance is a combination of different atoms?
    10·2 answers
  • Disadvantage of intrapreneurship​
    14·1 answer
  • Which of the following shows the correct order of structures? A. cells organs tissues systems B. cells systems tissues organs C.
    11·2 answers
  • the air pressure at the base of the mountain is 75.0cm of mercury while at the top is 60cm of mercury. Given that theaverage den
    5·1 answer
  • What provides all of the energy required to drive convection within the atmosphere and oceans?
    15·1 answer
  • Does a wheel barrow have mechanical advantage???
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!