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3 years ago

8

Technician A says that a lack of lubrication on the back of the disc brake pads can cause brake noise. Technician B says that pa

ds that are not correctly crimped to the caliper housing can cause brake noise. Which technician is correct?

1 answer:

sweet-ann [11.9K]3 years ago

7 0

Answer:

Technician A and B both are correct.

Explanation:

According to Technician A, we can stop noise by applying lubricant (Break friction).

According to Technician A, we can stop noise by set-up right position for break pad.

You might be interested in

List all possible fracture mechanisms under which the unidirectional composites fail. Briefly explain and describe the related m

professor190 [17]

Answer:

Ususushehehehhuuiiïbbb

Explanation:

Yyshehshehshshsheyysysueueue

7 0

2 years ago

A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Po

kicyunya [14]

This question is incomplete, the missing image in uploaded along this answer below.

Answer:

The required stress is 200 Mpa

Explanation:

Given the data in the question;

diameter D = 12 mm = 12 × 10⁻³ m

Length L = 188 mm = 188 × 10⁻³ m

Poisson's ratio v = 0.34

Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m

The transverse strain will;

εˣ = Δd / D

εˣ = -0.0105 × 10⁻³ / 12 × 10⁻³ m

εˣ = -0.00088

The longitudinal strain will be;

$E^z$ = - ( εˣ / v )

$E^z$ = - ( -0.00088 / 0.34 )

$E^z$ = - ( - 0.002588 )

$E^z$ = 0.0026

Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.

From the graph, in the Second image;

The stress is 200 Mpa

Therefore, The required stress is 200 Mpa

8 0

2 years ago

A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such t

bekas [8.4K]

Answer:

a)Δs = 834 mm

b)V=1122 mm/s

$a=450\ mm/s^2$

Explanation:

Given that

$s = 15t^3 - 3t\ mm$

a)

When t= 2 s

$s = 15t^3 - 3t\ mm$

$s = 15\times 2^3 - 3\times 2\ mm$

s= 114 mm

At t= 4 s

$s = 15t^3 - 3t\ mm$

$s = 15\times 4^3- 3\times 4\ mm$

s= 948 mm

So the displacement between 2 s to 4 s

Δs = 948 - 114 mm

Δs = 834 mm

b)

We know that velocity V

$V=\dfrac{ds}{dt}$

$\dfrac{ds}{dt}=45t^2-3$

At t= 5 s

$V=45t^2-3$

$V=45\times 5^2-3$

V=1122 mm/s

We know that acceleration a

$a=\dfrac{d^2s}{dt^2}$

$\dfrac{d^2s}{dt^2}=90t$

a= 90 t

a = 90 x 5

$a=450\ mm/s^2$

4 0

3 years ago

List two ways you can make an informal survey

solniwko [45]

The two ways you can use to make an informal survey are:

make field observations
interview people using informal unstructured techniques

<h3>What are informal surveys?</h3>

In informal surveys can be regarded as a type of survey that can be made by the researcher by going to the field themselves and this can be done by using different methods or ways.

For instance, the researcher can go out to interview people that can give the data that is needed about the research such as informally asking them questions, unstructured techniques can also be used to solve critical issues.

learn more about survey at: brainly.com/question/6947486

#SPJ9

8 0

2 years ago

Read 2 more answers

An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f

Gelneren [198K]

Answer:

(a) $Q=332$ $kvar$ and $C=5.66$ $uF$

(b) $pf=0.90$ lagging

Explanation:

Given Data:

$P=600kW$

$V=12.47kV$

$f=60Hz$

$pf_{old} =0.75$

$pf_{new} =0.95$

(a) Find the required kVAR rating of a capacitor

$\alpha _{old}=cos^{-1}(0.75) =41.41$ °

$\alpha _{new}=cos^{-1}(0.95) =18.19$ °

The required compensation reactive power can be found by

$Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))$

$Q=600(tan(41.41) - tan(18.19))$

$Q=332$ $kvar$

The corresponding capacitor value can be found by

$C=Q/2\pi fV^{2}$

$C=332/2*\pi *60*12.47^{2}$

$C=5.66$ $uF$

(b) calculate the resultant supply power factor

First convert the hp into kW

$P_{mech} =250*746=186.5$ $kW$

Find the electrical power (real power) of the motor

$P_{elec} =P_{mech}/n$

where $n$ is the efficiency of the motor

$P_{elec} =186.5/0.80=233.125$ $kW$

The current in the motor is

$I_{m} =(P/\*V*pf)$

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

$I_{m} =(233.125/12.47*0.85)$

$I_{m}=18.694+11.586$ $j$ $A$

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

$I_{load} =(600/12.47*0.75)$

$I_{load} =48.115-42.433$ $j$ $A$

Now the supply current is the current flowing in the load plus the current flowing in the motor

$I_{supply} =I_{m} + I_{load}$

$I_{supply}= (18.694+11.586)+(48.115-42.433)$

$I_{supply} =66.809-30.847j$ $A$

or in polar form

$I_{supply} =73.58$ °

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

$pf=cos(24.78)=0.90$ lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0

3 years ago