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kramer
3 years ago
8

Which of the following reduces friction in an engine A)wear B)drag C)motor oil D)defractionation

Engineering
2 answers:
kobusy [5.1K]3 years ago
3 0

It is motor oil, as oil is used to reduce friction

Ad libitum [116K]3 years ago
3 0
The answer would be motor oil
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A liquid refrigerant (sg=1,2) is flowing at a weight flow rate of 20,9 N/h. Refrigerant flashes into a vapor and its specific we
Iteru [2.4K]

Answer:

Explanation:

volume of 20.9 N

= 20.9 / 11.5 m³

= 1.8174 m³

In one hour 1.8174 m³ flows

in one second volume flowing = 1.8174 / 60 x 60

= 5 x 10⁻⁴ m³

Rate of volume flow = 5 x 10⁻⁴ m³ / s .

5 0
3 years ago
Differentiate between "Threshold and Resolution" with suitable examples.
9966 [12]

Answer:

to make the bace of a building more sturdy

Explanation:

example: the bace of the empire state building is stone very sturdy

6 0
2 years ago
(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1
olya-2409 [2.1K]

Answer:

The answer is "4.35 \times 10^{-3}\  mm and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point  

y =  yo + \frac{k}{\sqrt{x}}

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.  

Replacing the equation for each condition:  

y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k

People can get yo = 275 MPa with both equations and k= 15.5 Mpa mm^{\frac{1}{2}}.

To substitute the answer,  

310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\  mm

In point b:

The equation is \sigma y = \sigma 0 + k y d^{\frac{1}{2}}

equation is:

75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\

by putting the above value in the formula we get the \sigma y value that is= 157.5 MPa

5 0
2 years ago
Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r
My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

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
2 years ago
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