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Dahasolnce [82]
3 years ago
13

A front wheel drive vehicle with four wheel disc brakes is pulling to the left. Tech A says an external kink or internal restric

tion in the LF brake line will result in this condition. Tech B says to use a compression fitting to repair a section of brake line. Who is correct? Tech A Tech A Tech B Tech B Both Both Neither
Engineering
1 answer:
STALIN [3.7K]3 years ago
5 0

Answer:

Tech A is correct.          

Explanation:

A front-wheel-drive pulling to the left can result from several factors. One of them is definitely a faulty break.

A correct diagnosis linking the problem to the brakes is when there is an internal restriction and the pull is constant to one side and gets worse when the brakes are applied.

To confirm this, one would need to lift the vehicle and rotate each wheel by hand to check for excessive friction.

So the restriction may be caused by:

  • brake calipers that are sticky to the drum
  • too much brake fluid in the brake master cylinder - this prevents the caliper pistons from retracting when the brakes are released
  • misadjusted drum brakes and or parking brakes.

Cheers

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Info security:
il63 [147K]

Answer:

True

Explanation:

Dual home host - it is referred to as the firewall that is incorporated with two or more networks. out of these two networks, one is assigned to the internal network and the other is for the network. The main purpose of the dual-homed host is to ensure that no Internet protocol traffic is induced between both the network.

The most simple example of a dual-homed host is a computing motherboard that is provided with two network interfaces.

7 0
3 years ago
Consider a continuous-flow, indraft supersonic wind tunnel, which uses a vacuum pump to draw atmospheric air from outside of the
BigorU [14]
I believe it’s B but I’m not sure
3 0
3 years ago
Propane burns at an equivalence ratio (ER) of 0.6, determine actual air-fuel ratio. If excess air is 5%, what will be the actual
dimaraw [331]

Answer:

when 5% excess air is supplied, moles of air supplied/moles of fuel = 23.81\times 1.05 =25

Explanation:

Equivalence ratio = 0.6

Equivalence ratio = Actual air to fuel ratio (AAFR)/ stoichiometric air to fuel ratio SAFR

combustion reaction of propane is

C_3H_8+ 5O_2 ----->3CO_2+4H_2O

From above reaction,  1 mole of propane, from the reaction, 5  moles of oxygen required,  

we know that air contains 21% O_2 and 79% N_2,

Therefore, moles of air based on stoichiometry = \frac{5}{0.21} = 23.81

Theoretical air to fuel ratio = \frac{23.81}{1} = 23.81

Given\frac{AFR}{SFR} = 0.6

Actual Air Fuel Ratio = 23.81\times 0.6 = 14.3

when 5% excess air is supplied, moles of air supplied/moles of fuel = 23.81\times 1.05 =25

6 0
3 years ago
Problem 2 (a) A sinusoid with a frequency of 2Hz is applied to a sampler/zero-order hold combination. The sampling rate is 10Hz.
madreJ [45]

Answer & Explanation:

(a) Frequency of 2Hz is applied to a sampler/zero-order hold combination

The sampling rate is 10Hz

List of all the frequencies present in the output that are less than 50Hz.

Adding:

fs + fm = 10 + 2

= 12 Hz

2fs + fm = 2 * 10 + 2

= 22 Hz

3fs + fm = 3 * 10 + 2

= 32 Hz

4fs + fm = 4 * 10 + 2

= 42 Hz

Subtracting:

fs - fm = 10 - 2

= 8 Hz

2fs - fm = 2 * 10 - 2

= 18 Hz

3fs - fm = 3 * 10 - 2

= 28 Hz

4fs - fm = 4 * 10 - 2

= 38 Hz

5fs - fm = 5 * 10 - 2

= 48 Hz

(b) Frequency of 8Hz is applied to a sampler/zero-order hold combination

The sampling rate is 10Hz

List of all the frequencies present in the output that are less than 50Hz.

Adding:

fs + fm = 10 + 8

= 18 Hz

2fs + fm = 2 * 10 + 8

= 28 Hz

3fs + fm = 3 * 10 + 8

= 38 Hz

Subtracting:

fs - fm = 10 - 8

= 2 Hz

2fs - fm = 2 * 10 - 8

= 12 Hz

3fs - fm = 3 * 10 - 8

= 22 Hz

4fs - fm = 4 * 10 - 8

= 32 Hz

5fs - fm = 5 * 10 - 8

= 42 Hz

3 0
3 years ago
Is your mom the love of your life
hodyreva [135]
No but i love her
if she was the love of my life that would be weied
7 0
3 years ago
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