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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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A ball of mass 0.440 kg moving east (+x direction) with a speed of 3.80 m/s collides head-on with a 0.220-kg ball at rest. If th
avanturin [10]

Answer:

The speed and direction of each ball after the collision is  1.27 m/s to East direction and  5.07 m/s to East direction.

Explanation:

given information

m₁ = 0.440 kg

v₁ = 3.80 m/s

m₂ = 0.220 kg

v₂ = 0

collision is perfectly elastic

v₁ - v₂ = - (v₁'- v₂')

v₁ =  - (v₁'- v₂')

v₂' = v₁ + v₁'

according to momentum conservation energy

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

m₁v₁  = m₁v₁' + m₂(v₁ + v₁')

m₁v₁  =  m₁v₁' + m₂v₁ + m₂v₁'

m₁v₁ - m₂v₁ = m₁v₁' + m₂v₁'

v₁ (m₁ - m₂) = (m₁ + m₂) v₁'

v₁' = (m₁ - m₂)v₁ / (m₁ + m₂)

    = (0.440 - 0.220) (3.8) / (0.440 + 0.220)

     = 1.27 m/s to East direction

v₂' = v₁ + v₁'

    = 3.8 + 1.27

= 5.07 m/s to East direction

4 0
3 years ago
4
Blizzard [7]

Answer:

0.2 J

Explanation:

The pendulum forms a right triangle, with hypotenuse of 50 cm and base of 30 cm.  The height of this triangle can be found with Pythagorean theorem:

c² = a² + b²

(50 cm)² = a² + (30 cm)²

a = 40 cm

The height of the triangle is 40 cm.  The height of the pendulum when it is at the bottom is 50 cm.  So the end of the pendulum is lifted by 10 cm.  Assuming the mass is concentrated at the end of the pendulum, the potential energy is:

PE = mgh

PE = (0.200 kg) (9.8 N/kg) (0.10 m)

PE = 0.196 J

Rounding to one significant figure, the potential energy is 0.2 J.

6 0
4 years ago
Please help ASAP
Mandarinka [93]

Answer:

5cm by 4cm by 10cm = 200

200 / 10 = 20

20 :>

6 0
3 years ago
What are some of the major differences between the ancient Olympics and modern-day Olympics? List and describe at least two diff
umka21 [38]

Answer:

1

The ancient Olympic games only allowed people of Greek descent to participate. The Salt Lake City Olympics featured 2600 athletes from 77 countries. Only a few hundred athletes participated in the ancient games.

#2

Only men were allowed to compete in the ancient Greek games. Athletic training in ancient Greece was part of every free male citizen's education. The first women to compete in the Olympics were Marie Ohnier and Mme. Brohy. They participated in croquet games in the 1900 Olympics.

3 0
3 years ago
A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from
o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}

There is NO initial vertical velocity, so:

\large\boxed{\Delta d= \frac{1}{2}at^2}}

Rearrange to solve for time:

2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

t = \sqrt{\frac{2(60)}{(9.8)}} = 3.5 s

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

\large\boxed{d_x = vt}\\\\d_x = 10(3.5) = \karge\boxed{35 m}

6 0
3 years ago
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