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Marina86 [1]
4 years ago
7

Technician A says that applying too much braking force can cause tires to lose traction. Technician B says that ABS systems have

helped to reduce the number of vehicle accidents each year. Who is correct?
Engineering
1 answer:
Deffense [45]4 years ago
7 0

Answer:

technically both are correct but tech b is more correct because generally the older the vehicle the worse abs is newer vehicles have abs that no matter how hard you push on the brake peddle the wheels shouldnt lock up

Explanation:

You might be interested in
What is measurement in term of electrical engineering ​
laila [671]

Explanation:

voltage, current and resistance are the Volt [ V ], Ampere [ A ] and Ohm [ Ω ]

3 0
3 years ago
technician a says an out-of-round drum can cause a pulsating brake pedal. technician b says an out-of-round drum can cause the b
denis-greek [22]

According to the question of the pulsating brake pedal, both A and B are correct.

What causes brake pulsation?

Brake pulsation is mainly caused by warped rotors/brake discs. Excessive hard braking or quick stops, which can significantly overheat the discs, are the primary causes of deformed rotors. When the discs overheat, the composition of the metal disc material changes, resulting in imperfections in the metal's surface. Hotspots are noticeable irregularities. They appear as discoloured areas of the disc material, which are often bluish or blackish in appearance. The brake pedal is the pedal which you press with your foot to slow or stop a vehicle. When the driver presses the brake pedal, the system automatically delivers the appropriate pressure required to prevent colliding with the vehicle in front.

To learn more about brake pulsation
brainly.com/question/28779956

#SPj4

8 0
2 years ago
Given the potential field in cylindrical coordinates, V = 100/ (z2+1) rho cos φV, and point P at rho = 3m, φ = 60°, z = 2m, find
Margaret [11]

Answer:

V = 30 V

vector (E) = -10*a_p + 17.32*a_Q - 24*a_z

mag(E) = 31.24 V/m

dV / dN = 31.2 V /m

a_N = 0.32*a_p -0.55*a_Q + 0.77*a_z

p_v = 276 pC / m^3

Explanation:

Given:

- The Volt Potential in cylindrical coordinate system is given as:

                                     

- The point P is at p = 3 , Q = 60 , z = 2

Find:

values at P for:

a.) V

b.) E

c.) E

d.) dV/dN

e.) aN

f.) rhov in free space

Solution:

a)

Evaluate the Volt potential function at the given point P, by simply plugging the values of position P. The following:

                                     V = \frac{100*3*cos(60)}{2^2 + 1}\\\\V = \frac{150}{5} = 30 V      

b)

To compute the Electric field from Volt potential we have the following relation:

                                    E = - ∀.V

Where, ∀ is a del function which denoted:

                                   ∀ = \frac{d}{dp}.a_p + \frac{d}{p*dQ}*a_Q + \frac{d}{dz}*a_z

Hence,

                   E = \frac{100*cos(Q)}{z^2 + 1}.a_p+ \frac{100*sin(Q)}{z^2 + 1}.a_Q + \frac{-200*p*zcos(Q)}{(z^2 + 1)^2}.a_z

Plug in the values for point P:

                E = \frac{100*cos(60)}{5}.a_p+ \frac{100*sin(60)}{5}.a_Q + \frac{-200*3*2*cos(60)}{(5)^2}.a_z\\\\E = - 10*a_p +17.32 *a_Q-24*a_z

c)

The magnitude of the Electric Field is given by:

               E = √((-10)^2 + (17.32)^2 + (24)^2)

               E = √975.9824

               E = 31.241 N/C

d)

The dV/dN is the Field Strength E in normal to the surface with vector N given by:

              dV / dN = | - ∀.V |

              dV / dN = | E | = 31.2 N/C

e)

a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:

               a_N =  - vector(E) / (dV/dN)\\\\a_N =  10 /31.2 *a_p - 17.32/31.2 *a_Q + 24/31.2 *a_z\\\\a_N =  0.32 *a_p - 0.55 *a_Q + 0.77 *a_z

f)

The charge density in free space:

              p_v = E.∈_o

Where, ∈_o is the permittivity of free space = 8.85*10^-12

              p_v = 31.24.8.85*10^-12

              p_v = 276 pC / m^3

             

4 0
3 years ago
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I) A sag vertical curve is to be designed to join a 4% grade to a 2% grade. If the design
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