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blsea [12.9K]
1 year ago
10

Types of lubricants on the market include:

Engineering
1 answer:
zhenek [66]1 year ago
8 0

Answer:

1. Oil

2. Grease

3. Penetrating Lubricants

4. Dry Lubricants

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For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about ____
7nadin3 [17]

Answer:

One

For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about _one___

<h3>what is supported mounted?</h3>
  • A structure that holds up or serves as a foundation for something else. Support is a synonym for mounting.

To learn more about it, refer

to brainly.com/question/25689052

#SPJ4

4 0
2 years ago
In a hydraulic system, accumulator is a device that collects liquid and keeps the liquid under pressure.
Bumek [7]
The answer is: true
6 0
3 years ago
How do I get my son to do his work?
valentinak56 [21]

Answer:

Explanation:

Reward him if he does his homework/work

If he doesnt do his homework, take things that he loves off of him. and tell him if he does his homework/work he will get them things back

8 0
3 years ago
For a LED diode that has a= 632 nm, then the A1 is equal to:​
alexgriva [62]

Answer:

1.693242

Explanation:

The colors in the Light emitting diodes have been identified by wavelength which is measured in nano-meters. Wavelength is a function of LED chip material. The LED diode which has a = 632 then A1 will be 1.63242, this is calculated by 1 / 632. Wavelength are important for human eye sensitivity. The colors emitted from the LED will depend on the semiconductor material.

5 0
3 years ago
Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450 K, 350 kPa, and velocity of 3
viktelen [127]

Answer:

Given that

Mass flow rate ,m=2.3 kg/s

T₁=450 K

P₁=350 KPa

C₁=3 m/s

T₂=300 K

C₂=460 m/s

Cp=1.011 KJ/kg.k

For ideal gas

P V = m R T

P = ρ RT

\rho_1=\dfrac{P_1}{RT_1}

\rho_1=\dfrac{350}{0.287\times 450}

ρ₁=2.71 kg/m³

mass flow rate

m= ρ₁A₁C₁

2.3 = 2.71 x A₁ x 3

A₁=0.28 m²

Now from first law for open system

h_1+\dfrac{C_1^2}{200}+Q=h_2+\dfrac{C_2^2}{2000}

For ideal gas

Δh = CpΔT

by putting the values

1.011\times 450+\dfrac{3^2}{200}+Q=1.011\times 300+\dfrac{460^2}{2000}

Q=1.011\times 300+\dfrac{460^2}{2000}-\dfrac{3^2}{200}-1.011\times 450

Q= - 45.49 KJ/kg

Q =- m x 45.49 KW

Q= - 104.67 KW

Negative sign indicates that heat transfer from air to surrounding

4 0
3 years ago
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