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vredina [299]
4 years ago
9

Technician A says that the ABS electrohydraulic unit can be bled using bleeder screws and the manual method. Technician B says t

hat a scan tool is often required to bleed the ABS electrohydraulic unit. Which technician is correct?a. Technician A only
b. Technician B only
c. Both Technicians A and B
d. Neither Technician A nor B
Physics
1 answer:
Lana71 [14]4 years ago
6 0

Answer:

c. Both Technicians A and B                                    

Explanation:

ABS stands for anti lock braking system. It prevents the vehicle to skid while braking action in slippery surfaces and prevents lockup of the wheels.

The ABS ban be bled using a bleeder screw. Bleeding the brake means removing the air inside for a firm brake pedal. Thus to remove the air, open the bleeder screw of the ABS and allow to pass the air out while turning on the engine.

A fully featured scan tool is required to fully bleed the air in the ABS system.  The bi-directional control of the modulator can purge air bubbles out that can not be done with normal bleeding.

Thus both technician A and technician B are correct.

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Frequency, υ = 2.09 Hz

We know that time period, T = 1/υ

∴ T = 1 ÷ 2.09

⇒ T = 0.4784 seconds

5 0
4 years ago
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Explain two ways in which changes on the earth’s surface are connected to changes below the earth’s surface.
FinnZ [79.3K]

Answer:

The Earth's surfaces changes in slow processes, erosion and weathering, and some changes are due to rapid processes, such as: landslides, volcanic eruptions, tsunamis and earthquakes.

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4 years ago
Suppose that the strings on a violin are stretched with the same tension and each has the same length between its two fixed ends
lesya692 [45]

Answer:

0.00384 kg/m

Explanation:

The fundamental frequency of string waves is given by

f=\frac{1}{2L}\sqrt{\frac{F}{\mu}}

For some tension (F) and length (L)

f\propto\frac{1}{\mu}

Fundamental frequency of G string

f_G=196\ Hz

Fundamental frequency of E string

f_E=659.3\ Hz

Linear mass density of E string is

\mu_E=3.4\times 10^{-4}\ kg/m

So,

\frac{F_G}{F_E}=\sqrt{\frac{\mu_E}{\mu_G}}\\\Rightarrow \frac{F_G^2}{F_E^2}=\frac{\mu_E}{\mu_G}\\\Rightarrow \mu_G=3.4\times 10^{-4}\times \frac{659.3^2}{196^2}\\\Rightarrow \mu_G=0.00384\ kg/m

The linear density of the G string is 0.00384 kg/m

4 0
3 years ago
Which element has 7 valence electrons ? A.nitogen(N) B.neon(Ne) C.lithium(Li) D.chlorine(CI)
aalyn [17]

D is the correct answer

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3 years ago
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What is the acceleration of a ball rolling down a ramp that starts from rest and travels 0.9 m in 3 s?
cupoosta [38]
Given:
u = 0, initial velocity
s 0.9 m, distance traveled.
t = 3 s, the time taken.

Let a =  the acceleration. Then
s = ut + (1/2)*a*t²
(0.9 m) = 0.5*(a m/s²)*(3 s)²
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Answer: 0.2 m/s²
3 0
3 years ago
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