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AfilCa [17]
3 years ago
14

The brakes are being bled on a passenger vehicle with a disc/drum brake system. Technician A says that the drums should be remov

ed before bleeding. Technician B says that the fluid in the master cylinder should be kept low during the bleeding procedure. Who is right?
Engineering
1 answer:
exis [7]3 years ago
4 0

The brakes are being bled on a passenger vehicle with a disc/drum brake system is described in the following

Explanation:

1.Risk: Continued operation at or below Rotor Minimum Thickness can lead to Brake system failure. As the rotor reaches its minimum thickness, the braking distance increases, sometimes up to 4 meters. A brake system is designed to take kinetic energy and transfer it into heat energy.

2.Since the piston needs to be pushed back into the caliper in order to fit over the new pads, I do open the bleeder screw when pushing the piston back in. This does help prevent debris from traveling back through the system and contaminating the ABS sensors

3.There are three methods of bleeding brakes: Vacuum pumping. Pressure pumping. Pump and hold.

4,Brake drag is caused by the brake pads or shoes not releasing completely when the brake pedal is released. ... A worn or corroded master cylinder bore causes excess pedal effort resulting in dragging brakes. Brake Lines and Hoses: There may be pressure trapped in the brake line or hose after the pedal has been released.

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Water from a stationary nozzle impinges on a moving vane with turning angle θ = 120. The vane moves away from the nozzle with co
Tems11 [23]

Answer:

The force that must be applied to maintain the vane speed constant is 2771.26 N

Explanation:

Given;

turning angle of the vane = 120°

control volume velocity, u = 10 m/s

absolute velocity, v = 30 m/s

nozzle area = 0.004 m²

The force acting on the vane has horizontal and vertical components:

Based on Reynolds general control volume system;

The horizontal force component of the system, ∑Fₓ = ρW²A(1-cosθ)

where;

ρ is the density of water = 1000 kg/m³

W is the relative velocity = Absolute velocity - control volume velocity

W = v - u

    = 30 - 10 = 20m/s

∑Fₓ = ρW²A(1-cosθ) = 1000 x 20² x 0.004 (1 - cos 120) = 2400 N

The vertical force component of the system, ∑Fy = ρW²A(sinθ)

∑Fy = ρW²A(sinθ) = 1000 x 20² x 0.004 x sin(120) = 1385.6 N

The magnitude of the force applied = \sqrt{F_x^2 + F_y^2}

F = \sqrt{2400^2 + 1385.6^2} = 2771.26 \ N

The force that must be applied to maintain the vane speed constant is 2771.26 N

6 0
3 years ago
Read 2 more answers
Calculate the areas under the stress-strain curve (toughness) for the materials shown in Fig. below, (a) plot them as a
defon

Answer:

Explanation:

Wow

5 0
3 years ago
A divided multilane highway in a recreational area has four lanes (two lanes in each direction) and is on rolling terrain. The h
sertanlavr [38]

Answer:

Explanation:

Before: PT= 0.10, PB= 0.03 (given) ET = 2.5 ER = 2.0 (Table 6.5)

fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)

vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)

BFFS = 50+5, BFFS =55 (given) fLW= 6.6

TLC=6+3=9 fLC= 0.65

fM= 0.0

fA= 1.0

FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)

Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)

After: fA= 3.0

FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)

Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln

8 0
3 years ago
The north-south streets of a grid have block lengths of 250 m and the east-west streets have block lengths of 200 m. Desired spe
antoniya [11.8K]

Answer:

See attached pictures.

Explanation:

See attached pictures for detailed explanation.

7 0
3 years ago
A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring
ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

5 0
3 years ago
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