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Masja [62]
3 years ago
9

Technician A says that the most commonly used combustion chamber types include hemispherical, and wedge. Technician B says that

coolant and lubricating openings and passages are located throughout most cylinder heads. Which technician is correct
Engineering
1 answer:
Inessa05 [86]3 years ago
6 0

Answer:

Technician A and Technician B both are correct.

Explanation:

Technician A accurately notes that perhaps the forms of combustion process most widely used are hemispherical and cross.

Technician B also correctly notes that in several cylinder heads, cooling system and greases gaps and pathways are found.

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A 3-m wide rectangular channel has a flow velocity of 1.8 m/s when the depth of flow is 1.2 m. what will be the flow velocity wh
valina [46]

Answer:

The flow velocity reduces to 0.72 m/s

Explanation:

According to the equation of continuity discharge in the channel should remain same

Thus we have

A_{1}V_{1}=A_{2}V_{2}

For a rectangular channel we have Area=Depth\times Width

Applying values in the continuity equation and since the width of the channel remains constant 3.0 m we have

3\times 1.8\times 1.2=3\times 3\times V_{2}\\\\\therefore V_{2}=\frac{6.48}{9}=0.72m/s

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Answer:

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Explanation:

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zubka84 [21]

Answer:

Routine

Explanation:

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While there are many ways to solve this problem, one strategy is to calculate the volume of any metal's unit cell given its theo
IgorC [24]

Answer:

V_{c}=1.396x10^{-28}  m^{3} /unit.cell

Explanation:

z = number of atoms

M = Molar mass of zirconium

N = Avogadro’s number

Vc = volume of zirconium unit cell

d = density

z=12x\frac{1}{6}+2x\frac{1}{2}+3=6

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M = 91.224 g/mol

N = 6.023x10^{23}  atoms/mol

d = 6.51g/cm^{3}

V_{c}=\frac{zxM}{dxN}

V_{c}=\frac{6x(91.224g/mol)}{(6.51g/cm^{3}) x(6.023x10^{23}atoms/mol) }

V_{c}=1.396x10^{-22}  cm^{3} /unit.cell

V_{c}=1.396x10^{-28}  m^{3} /unit.cell

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