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bulgar [2K]
3 years ago
12

All of the following are drum brake components mounted to the backing plate, EXCEPT:

Engineering
1 answer:
AlexFokin [52]3 years ago
5 0

Answer:

B. wheel cylinders

Explanation:

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The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8
il63 [147K]

Answer:

a.)

US Sieve no.                         % finer (C₅ )

4                                                  100

10                                                95.61

20                                               82.98

40                                               61.50

60                                               42.08

100                                              20.19

200                                              6.3

Pan                                               0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no.             Mass of soil retained (C₂ )

4                                            0

10                                          18.5

20                                         53.2

40                                         90.5

60                                         81.8

100                                        92.2

200                                       58.5

Pan                                        26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×\frac{100}{w}

∴ we get

US Sieve no.               % retained (C₃ )               Cummulative % retained (C₄)

4                                            0                                           0

10                                          4.39                                      4.39

20                                         12.63                                     17.02

40                                         21.48                                     38.50

60                                         19.42                                     57.92

100                                        21.89                                     79.81

200                                       13.89                                     93.70

Pan                                        6.30                                      100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no.               Cummulative % retained (C₄)          % finer (C₅ )

4                                                     0                                          100

10                                                  4.39                                      95.61

20                                                 17.02                                     82.98

40                                                 38.50                                    61.50

60                                                 57.92                                    42.08

100                                                79.81                                     20.19

200                                                93.70                                   6.3

Pan                                                 100                                        0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = \frac{D60}{D10}

⇒ Cu = \frac{0.4}{0.12} = 3.33

d.)

Coefficient of Graduation = Cc = \frac{D30^{2}}{D10 . D60}

⇒ Cc = \frac{0.22^{2}}{(0.4) . (0.12)} = 1

3 0
3 years ago
1. When the pneumatic device was demonstrated, explain something that was different than what you expected.
Vera_Pavlovna [14]

Answer:

1. Effect of air pressure

2. air- powered wheel chair

3. Pneumatic valves

Explanation:

1. In any pneumatic device, the mipact of air pressure to produce the moving effect on an heavy object is unexpected.

2. pneumatice demultiplexer when air in comprressed tank is allowed released to cause movement of the  chair.

3. In industries, a pneumatic valve operates by force of air when actuated. A signal causes actuation of coil. When coil is energized, compressed high pressure air is allowe to enter in a small cylinder and cause operation of valve

5 0
3 years ago
Read 2 more answers
Describe why the motion of a follower acted on by a cam is periodic motion.
PolarNik [594]
When you go and use a periodic motion there a reason
6 0
3 years ago
An injector pressure drop test to see if the injector is restricted with deposits can be done using an?
Vitek1552 [10]

An injector pressure drop test to see if the injector is restricted with deposits can be done using an integer test.

<h3>What is injector strain drop?</h3>

Fuel injectors have a glide charge. This glide charge of the injector is rated at a sure strain drop throughout the injector. Meaning the injector glide is say 30#/hr at 42.five PSI. This approach that that injector will glide 30#/hr so long as the strain on the deliver side, minus the strain withinside the manifold is 42.five psiListening or Clicking Test.

Start the engine and permit it to idle. Keep the engine strolling and contact the give-up of a protracted steel screwdriver towards the gas injector. Put your ear on the alternative give up of the screwdriver. A clicking sound approaches the injector's working.

Read more about the injector pressure drop:

brainly.com/question/12385665

#SPJ1

6 0
2 years ago
A CSTR is being used to carry out a certain reaction isothermally in the liquid phase. It is observed that the conversion in the
Semmy [17]

Answer:

The high conversion in a CSTR with not very well mixing occurs mostly if the reactor is micro reactor.

From Literature, mixing due to diffusion depends on concentration or temperature gradients. This is commonly represented in micro reactors.

Also, the residence time distribution (RTD) Analysis provides information on how long the various components of fluid have been in the reactor which leads to more conversion in some cases of fluid reaction.

8 0
3 years ago
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