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WARRIOR [948]
3 years ago
14

Technician A says that the traction control system was developed to prevent the drive wheels from slipping while the vehicle is

being decelerated when coming to a stop. Technician B says that above a manufacturer-specified speed, the traction control system deactivates. Who is correct
A. Technician A Only
B. Technician B Only
C. Bpth A and B
D. Neither A nor B
Physics
2 answers:
dmitriy555 [2]3 years ago
8 0

Answer:

The correct option is;

B. Technician B Only

Explanation:

The traction control system (TCS) also known as ASR or drive slippage regulation is a safety feature of the vehicle stability control present in automobiles. TCS is brought into play at the instant there is a mismatch between the throttling and the torque of the engine to the condition of the surface of the road.

TCS is functional to a speed specified by the automobile manufacturer above which the Electronic Brake Control Module (EBCM) deactivates the wheels will not loose traction when the acceleration is further increased

An Anti-lock Braking System (ABS) is meant to prevent skidding.

mihalych1998 [28]3 years ago
3 0

Answer:

Option B.

Technician B Only is correct

Explanation:

Traction Control systems are systems in a vehicle that are deigned to  optimize grip and stability of the car on the road during acceleration by measuring wheel rotation.

Traction systems are designed to work during acceleration, not deceleration. this makes Technician A wrong.

Traction control systems are pretty much  active up to a manufacturer-specified speed. Above that speed, the  traction control is deactivated by the electronic brake control module because further acceleration is unlikely to cause the wheels to lose traction. This makes Technician B correct

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natali 33 [55]
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3 years ago
un movil que parte del reposo alcanza una velocidad de 75 m/s en 13 segundos ¿cual su aceleracion y el espacio que recorrio en l
Dmitriy789 [7]

Answer:

Acceleration = 5.77 m/s²

Distance cover in 13 seconds = 487.56 meter

Explanation:

Given:

Final velocity of mobile device = 75 m/s

initial velocity of mobile device = 0 m/s

Time taken = 13 seconds

Find:

Acceleration

Distance cover in 13 seconds

Computation:

v = u + at

75 = 0 + (a)(13)

13a = 75

a = 5.77

Acceleration = 5.77 m/s²

s = ut + (1/2)(a)(t²)

s = (0)(t) + (1/2)(5.77)(13²)

Distance cover in 13 seconds = 487.56 meter

8 0
3 years ago
two horses pull against a rope with forces of 100 newtons in opposite directions. this is an example of
statuscvo [17]
It is an example of balanced force.


hope this helps. good luck 
4 0
3 years ago
Read 2 more answers
si se deja caer un carrito desde el punto mas alto de ua psta de coches cuya altura es de 1.4m cual es la velocidad maxima que p
forsale [732]

Answer:

v = 5.24[m/s]

Explanation:

Este problema se puede resolver por medio del principio de la conservación de la energía, donde la energía potencial es igual a la energía cinética. Es decir a medida que el carrito desciende su energía potencial disminuye, pero su energía cinética aumenta.

E_{kin}=E_{pot}

Donde:

E_{kin}=\frac{1}{2} *m*v^{2} \\\\E_{pot}=m*g*h

Ahora reemplazando:

\frac{1}{2} *m*v^{2}=m*g*h\\\\0.5*v^{2}=9.81*1.4\\v=\sqrt{\frac{9.81*1.4}{0.5} }   \\\\v=5.24[m/s]

6 0
3 years ago
A 250KG ROCKET HAS 1,953,125 JOULES OF ENERGY. HOW FAST IS IT<br>MOVING?​
UNO [17]

Answer:

cnbv gncvjhmv

Explanation:

6 0
4 years ago
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