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3 years ago

In multi-grade oil what is W means?

Engineering

1 answer:

irga5000 [103]3 years ago

4 0

Answer:

winter viscosity grades

Explanation:

The “W”/winter viscosity grades describe the oil's viscosity under cold temperature engine starting conditions. There's a Low Temperature Cranking Viscosity which sets a viscosity requirement at various low temperatures to ensure that the oil isn't too thick so that the starter motor can't crank the engine over.

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Twenty-five wooden beams were ordered or a construction project. The sample mean and he sample standard deviation were measured

aksik [14]

Answer:

Correct option: B. 90%

Explanation:

The confidence interval is given by:

$CI = [\bar{x} - z\sigma_{\bar{x}} , \bar{x}+z\sigma_{\bar{x}} ]$

If $\bar{x}$ is 190, we can find the value of $z\sigma_{\bar{x}}$ :

$\bar{x} - z\sigma_{\bar{x}} = 188.29$

$190 - z\sigma_{\bar{x}} = 188.29$

$z\sigma_{\bar{x}} = 1.71$

Now we need to find the value of $\sigma_{\bar{x}}$ :

$\sigma_{\bar{x}} = s / \sqrt{n}$

$\sigma_{\bar{x}} = 5/ \sqrt{25}$

$\sigma_{\bar{x}} = 1$

So the value of z is 1.71.

Looking at the z-table, the z value that gives a z-score of 1.71 is 0.0436

This value will occur in both sides of the normal curve, so the confidence level is:

$CI = 1 - 2*0.0436 = 0.9128 = 91.28\%$

The nearest CI in the options is 90%, so the correct option is B.

4 0

3 years ago

Consider tests of an unswept wing that spans the wind tunnel and whose airfoil section is NACA 23012. Since the wing model spans

Dominik [7]

Answer:

Check the explanation

Explanation:

to know the lift per unit span (N/m) that is expected to be measured when the wing attack angle is 4°

as well as the corresponding section lift coefficient and die moment coefficient .

Kindly check the attached image below to see the step by step explanation to the above question.

3 0

3 years ago

The toggle (t) flip-flop has one input, clk, and one output, q. on each rising edge of clk, q toggles to the complement of its p

jeyben [28]

Answer:

See attached

Explanation:

The next state of a toggle flip-flop is the inverse of the present state. This behavior can be produced using a D flip-flop that has its input connected to the inverse of its output.

A schematic is attached.

7 0

2 years ago

List and describe three classifications of burns to the body.

DiKsa [7]

AnswerWhat Are the Classifications of Burns? Burns are classified as first-, second-, or third-degree, depending on how deep and severe they penetrate the skin's surface. First-degree burns affect only the epidermis, or outer layer of skin. The burn site is red, painful, dry, and with no blisters.

Explanation:

8 0

3 years ago

Read 2 more answers

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago