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alex41 [277]
3 years ago
11

11. Which of the following is the brake fluid most often used?

Engineering
2 answers:
Olenka [21]3 years ago
4 0
Dot 3 is mostly used in a lot of v4 and v6
Nadya [2.5K]3 years ago
4 0

Answer:

B. DOT 3

Explanation:

The braking system of a vehicle uses hydraulic power generated by a master cylinder to activate the four-wheel brake assemblies.  The hydraulic power is transmitted throughout the brake system via brake fluid.  Brake fluid is made from  a silicone based or glycol based product.  .The three classes of brake fluid commonly used in modern automobiles are as follows:

DOT 3, DOT 4, and DOT 5. DOT stands for

Department of Transportation specification grades; the increasing numbers corresponds to increasing boiling point. DOT 3 and DOT 4 are glycol based, and  DOT 3 most commonly used in disc brake systems. The glycol base is hygroscopic means it attract water from the environment.

brake fluid should be disposed properly. it should not be exposed to fire as this mit pose a ue fire hazard. t shouldn't be thrown into septic tanks.

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7.35 and 7.36 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute
Crank

Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

<h3>How to draw Shear Force and Bending Moment Diagram?</h3>

A) We can see the beam loaded in the first image attached.

For the shear diagram, let us calculate the shear from point load to point load.

From A to C, summing vertical to zero gives; ∑fy = 0: -20 - V = 0

V = -20 KN

From C to D, summing vertical to zero gives; ∑fy = 0: -20 + 48 - V = 0

V = 28 KN

From D to E, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - V = 0

V = 8 KN

From E to B, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - 20 - V = 0

V = -12 KN

For the bending moment diagram, let us calculate the bending moment from point load to point load.

At point A, the bending moment would be zero. Thus, M_A = 0 KN.m

At point C, taking moment about point C and equating to zero gives;

M_C = 0. Thus; 20(0.225) + M = 0

M = -4.5 KN.m

At point D, taking moment about point D and equating to zero gives;

M_D = 0. Thus; 20(0.525) - 48(0.3) + M = 0

M = 3.9 KN.m

At point E, taking moment about point E and equating to zero gives;

M_D = 0. Thus; 20(0.75) - 48(0.525) + 20(0.225) + M = 0

M = 5.7 KN.m

At point B, taking moment about point E and equating to zero gives;

M_E = 0. Thus; 20(1.05) - 48(0.825) + 20(0.525) + (20 * 0.3) + M = 0

M = 2.1 KN.m

2) From the attached diagrams, we can deduce that;

Maximum absolute values of the shear = 28 KN

Maximum absolute values of bending moment = 5.7 KN.m

Read more about shear force & bending moment diagram at; brainly.com/question/14834487

#SPJ1

4 0
1 year ago
If a front gear had 24 teeth, and a rear gear has 12 teeth:
zubka84 [21]

Answer:

  4 times around

Explanation:

The total number of teeth involved will be the same for each gear. If the front gear is connected to the pedal and it goes around twice, then 2·24 = 48 teeth will have passed the reference point.

If the rear gear is attached to the wheel, and 48 teeth pass the reference point, then it will have made ...

  (48 teeth)/(12 teeth/turn) = 4 turns

4 0
2 years ago
A tank with a volume of 8 m3 containing 4 m3 of 20% (by volume) NaOH solution is to be purged by adding pure water at a rate of
lawyer [7]

Answer:

The time necessary to purge 95% of the NaOH is 0.38 h

Explanation:

Given:

vfpure water(i) = 3 m³/h

vNaOH = 4 m³

xNaOH = 0.2

vfpure water(f) = 2 m³/h

pwater = 1000 kg/m³

pNaOH = 1220 kg/m³

The mass flow rate of the water is = 3 * 1000 = 3000 kg/h

The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg

When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg

The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³

The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h

4 0
3 years ago
The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod
Vsevolod [243]

Answer:

Explanation:

From the information given:

E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa  \\ \\ d = 25 \ mm \  \\ \\ D = 45 \ mm \ \\ \\ L   = 761 \ mm  \\ \\ P = -88 kN

The total load is distributed across both the rod and tube:

P = P_1+P_2 --- (1)

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

\delta_1=\delta_2

\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}

\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}

P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})

P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}

P_2 = 6.6212 \ P_1

Replace P_2 into equation (1)

P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\  -88 = 7.6212 \ P_1  \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\  P_1 = -11.547 \ kN

Finally, to determine the normal stress in aluminum rod:

\sigma _1 = \dfrac{P_1}{A_1} \\ \\  \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}

\sigma_1 = - 23.523 \ MPa}

Thus, the normal stress = 23.523 MPa in compression.

8 0
3 years ago
g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0
3 years ago
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