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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.

Explanation:

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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Rotary compressors are used primarily in ____. A. large industrial refrigeration and air conditioning systems B. high volume res

RideAnS [48]

Answer:

Answer is D.window units / small residential units

Refer below.

Explanation:

Rotary compressors are used primarily in window units / small residential units

7 0

3 years ago

Two identical circular, wire loops 35.0 cm in diameter each carry a current of 2.80 A in the same direction. These loops are par

defon

Answer:

The answer is " $4659.2 \times 10^{-24} \ N$ "

Explanation:

The magnetic field at ehe mid point of the coils is,

$\to B=\frac{\mu_0 i R^2}{(R^2+x^2)^{\frac{3}{2}}}\\\\$

Here, i is the current through the loop, R is the radius of the loop and x is the distance of the midpoint from the loop.

$\to B=\frac{(4\pi\times 10^{-7})(2.80\ A) (\frac{0.35}{2})^2}{( (\frac{0.35}{2})^2+ (\frac{0.24}{2})^2)^{\frac{3}{2}}}\\\\$

$=\frac{(12.56 \times 10^{-7})(2.80\ A) \times 0.030625}{( 0.030625+ 0.0144)^{\frac{3}{2}}}\\\\=\frac{ 1.07702 \times 10^{-7} }{0.0095538976}\\\\=112.730955 \times 10^{-7}\\\\=1.12\times 10^{-5}\ \ T\\$

Calculating the force experienced through the protons:

$F=qvB=(1.6 \times 10^{-19}) (2600)(1.12 \times 10^{-5})= 4659.2 \times 10^{-24}\ N$

7 0

3 years ago

As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury baromet

Pani-rosa [81]

Answer:

new atmospheric pressure is 0.9838 × $10^{5}$ Pa

Explanation:

given data

height = 21.6 mm = 0.0216 m

Normal atmospheric pressure = 1.013 ✕ 10^5 Pa

density of mercury = 13.6 g/cm³

to find out

atmospheric pressure

solution

we find first height of mercury when normal pressure that is

pressure p = ρ×g×h

put here value

1.013 × $10^{5}$ = 13.6 × 10³ × 9.81 × h

h = 0.759 m

so change in height Δh = 0.759 - 0.0216

new height H = 0.7374 m

so new pressure = ρ×g×H

put here value

new pressure = 13.6 × 10³ × 9.81 × 0.7374

atmospheric pressure = 98380.9584

so new atmospheric pressure is 0.9838 × $10^{5}$ Pa

6 0

3 years ago

Explain how you can improve the accuracy of a measurement.

yKpoI14uk [10]

You use more significant figures. 5 sigfigs (1.0985) is more accurate than 2 sigfigs (1.0)

8 0

2 years ago

Read 2 more answers

A skydiver jumps out of a plane flying at an altitude of 5,500 meters. How fast is the skydiver going when they get to the groun

andrey2020 [161]

Answer:

2,500 feet (760 meters)

Explannation: At about 2,500 feet (760 meters), the skydiver throws out a pilot chute, and it deploys the parachute. Its used to control the fall rate.

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2 years ago