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Keith_Richards [23]
3 years ago
15

Technician A says the parking brake cable adjustment should be performed before adjusting the rear brakes. Technician B says the

parking brake cable should allow for about 15 "clicks" before the parking brake holds. Which technician is correct?
Physics
1 answer:
Firlakuza [10]3 years ago
7 0

Answer:

Neither both are correct

Explanation:

The question involves the removal and installation procedure of cable for parking break. The answer for this could be found through the manual procedure for repairing parking break

Technician A is wrong because rear brakes should be adjusted before the parking cable adjustment is being made.

Technician B is wrong because the parking brake lever should be intact and secure at all clicks to allow maximum security. 15 clicks in this case should be the maximum number of clicks for the lever.

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How much work (in J) is required to expand the volume of a pump from 0.0 L to 2.7 L against an external pressure of 1.0 atm?
spin [16.1K]

Answer:

-2.79 x 10²J

Explanation:

using the pressure volume work formula which states that

work = -PΔV

           = -(1.0 atm) (2.7-0.0)L = -2.7L . atm

Convert litre.atmosphere to Joules.

1 L . atm = 101.325 joules

-2.75 L .atm = -2.75 x 101.325 = -278.64375 =

Work = -2.79 x 10²J

5 0
4 years ago
A horizontal beam of electrons initially moving at 4.0×10^7 m/s is deflected vertically by the vertical electric field between o
givi [52]

Answer:

1.77\times 10^{-7}\ C/m^2

0.000439077936334 m

Explanation:

q = Charge of electron = 1.6\times 10^{-19}\ C

E = Electric field = 2\times 10^{4}\ N/C

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

d = Distance between plates = 2 cm (assumed)

m = Mass of electron = 9.11\times 10^{-31}\ kg

The beam consists of electrons which means it has negative charge this means the upper plates will be positive and the lower plate will be negative.

The direction is upper to lower lower plate.

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

Electric flux is given by

\phi=\epsilon_0E\\\Rightarrow \phi=8.85\times 10^{-12}\times 2\times 10^{4}\\\Rightarrow \phi=1.77\times 10^{-7}\ C/m^2

The charge per unit area on the plates is 1.77\times 10^{-7}\ C/m^2

Deflection is given by

s=\dfrac{1}{2}\dfrac{qE}{m}(\dfrac{d}{v})^2\\\Rightarrow s=\dfrac{1}{2}\dfrac{1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}(\dfrac{0.02}{4\times 10^7})^2\\\Rightarrow s=0.000439077936334\ m

The deflection is 0.000439077936334 m

7 0
3 years ago
A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth
Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

\tan\theta=\dfrac{8}{3}

\thta=\tan^{-1}\dfrac{8}{3}

\theta=69.44^{\circ}

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

mg=2T\cos\theta

Put the value into the formula

3\times9.8=2T\times\cos69.44

T=\dfrac{3\times9.8}{2\times\cos69.44}

T=41.85\ N  

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

4 0
4 years ago
Read 2 more answers
Ohm's Law for electrical circuits is stated Vequals​RI, where V is a constant​ voltage, R is the resistance in ohms and I is the
Annette [7]

Answer:

The resistance interval is  R  =  1.8 \pm  0.037

Explanation:

From the question we are told that

     The voltage is  V  =  9 V

      The current is  I = 5 \pm 0.1

The maximum current would be  

       I_{max} = 5 + 0.1 =  5.1 \ A

The minimum current would be  

      I_{min} = 5 -  0.1 =  4.9 \ A

The maximum resistance is  

      R_max =  \frac{V}{I_{min}}

     R_max =  \frac{9}{4.9}

     R_max =  1.837 \Omega

The minimum resistance is  

      R_{min} =  \frac{V}{I_{max}}

    R_{min} =  \frac{9}{5.1}

    R_{min} = 1.765 \Omega

and  R  =  \frac{9}{5}  =  1.8 \Omega

The  interval R  lies is  

        R  =  1.8 \pm  0.037

4 0
3 years ago
What force is affected when the distance between two objects remains the same and the mass of each object is doubled?.
rusak2 [61]

Answer:

Gravitational force between masses there is Gravitational force .

3 0
3 years ago
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