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zepelin [54]
3 years ago
6

What are wheel cylinders used for?

Engineering
1 answer:
8090 [49]3 years ago
4 0

Answer:

A hydraulic brake system

Explanation:

A wheel cylinder is a component of a hydraulic brake system. It is located in each wheel and is usually positioned at the top of the wheel, above the shoes. Its function is to exert force onto the shoes so as to bring them into contact with the drum and stop the vehicle with friction.

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The peripheral system allows the brain and spinal cord to receive and send information to other areas of the body, which allows us to react to stimuli in our environment.

The nerves that make up the peripheral nervous system are actually the axons or bundles of axons from nerve cells or neurons.

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Guess my birthday it’s in September you should be good with it
Paraphin [41]

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spetember 7th

Explanation:

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3 years ago
What is structural analysis
Luden [163]
DescriptionStructural analysis is the determination of the effects of loads on physical structures and their components.
8 0
4 years ago
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The transfer function of a typical tape-drive system is given by
maw [93]

Answer:

the range of K can be said to be :  -3.59 < K< 0.35

Explanation:

The transfer function of a typical tape-drive system is given by;

KG(s) = \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]}

calculating the characteristics equation; we have:

1 + KG(s) = 0

1+   \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]} = 0

{s[s+0.5)(s+1)(s^2+0.4s+4)]} +{K(s+4)}= 0

s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ 2s+K(s+4) = 0

s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ (2+K)s+ 4K = 0

We can compute a Simulation Table for the Routh–Hurwitz stability criterion Table as  follows:

S^5             1                     5.1                          2+ K

S^4            1.9                   6.2                           4K

S^3             1.83            \dfrac{1.9 (2+K)-4K}{1.9}          0

S^2        \dfrac{11.34-1.9(X)}{1.83}       4K                         0

S          \dfrac{XY-7.32 \ K}{Y}        0                            0

\dfrac{1.9 (2+K)-4K}{1.9} = X

 

\dfrac{11.34-1.9(X)}{1.83}= Y

We need to understand that in a given stable system; all the elements in the first column is usually greater than zero

So;

11.34  - 1.9(X) > 0

11.34  - 1.9(\dfrac{3.8+1.9K-4K}{1.9}) > 0

11.34  - (3.8 - 2.1K)>0

7.54 +2.1 K > 0

2.1 K > - 7.54

K > - 7.54/2.1

K > - 3.59

Also

4K >0

K > 0/4

K > 0

Similarly;

XY - 7.32 K > 0

(\dfrac{3.8+1.9K-4K}{1.9})[11.34  - 1.9(\dfrac{3.8+1.9K-4K}{1.83}) > 7.32 \ K]

0.54(2.1K+7.54)>7.32 K

11.45 K < 4.07

K < 4.07/11.45

K < 0.35

Thus the range of K can be said to be :  -3.59 < K< 0.35

4 0
3 years ago
dentify a semiconducting material and provide the value of its band gap) that could be used in: (a) (1 point) red LED (b) (1 poi
OLga [1]

Answer:

(a) Aluminum Indium Gallium Phosphide (AlInGaP). Band gap = 1.81eV  ≈ 2eV

(b) Gallium Nitride (GaN). Band Gap = 3.4eV

(c) Aluminium Gallium Arsenide (AlGaA). Band Gap = 1.42eV ≈ 2.16eV

(d) Zinc Selenide (ZnSe). Band Gap = 2.82eV

(e) Gallium Phosphide (GaP). Band Gap = 2.24eV

Explanation:

LED's are semi-conducting materials that convert electrical energy to light energy. The light color emitted from the LED depends on the semi-conducting material and other compositions.

The band gap of the semi-conductor determines its wavelength. High band gap semi-conductors emit lower wavelengths which means greater power(UV semi-conducting macterials fall under this category).

5 0
3 years ago
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