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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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Positively charged ions are surrounded by freely moving electrons in

ASHA 777 [7]

I believe the answer would be valence shells

3 0

3 years ago

A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas

konstantin123 [22]

Answer:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Explanation:

The expression for conservation of the angular momentum (L) is

$L_{i} = L_{f} I_{i}\times\omega_{i} = I_{f}\times\omega_{f}$

Where

$I_{i}\ and \ \omega_{i}$ initial moment of inertia and angular velocity

$I_{f}\ and \ \omega_{f}$ is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

$I_{i} = \frac{2}{5}m_{s}r^{2}$

Where $m_{s}$ is the satellite mass

r is the radus of the sphere

Substititute 1900kg for m and 4.6m for r

$I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}$

The final moment of inertia of the satellite about the centre of mass

$I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}$

Where $m_{x}$ is the antenna's mass and

I is the length of the antenna

$I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}$

So, the Final rotation rate of the satellite is:

$I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s$

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

5 0

3 years ago

The two spheres pictured above have equal densities and are subject only to their mutual gravitational attraction. Which of the

kiruha [24]

Answer:

Gravitational force

Explanation:

If two spheres have equal densities and they are subject only to their mutual gravitational attraction. We need to say that the quantities that must have the same magnitude for both spheres. So, the correct option is (E) i.e. gravitational force.

It is because of Newton's third law of motion. It states that the force due to object 1 to object 2 is same as force due to object 2 to object 1. The two forces act in opposite direction.

Hence, the correct option is (E) "Gravitational force".

8 0

2 years ago

When an oxygen atom forms an ion, it gains two electrons

kirill [66]

Note the atom of the Oxygen is electrically neutral, meaning it has equal numbers of electrons and protons.

So if it gains 2 electrons, it would have excess of 2 electrons, hence its charge would be -2.

Option B.

5 0

2 years ago

Pls help on this one?

sattari [20]

The answer is point C

4 0

3 years ago