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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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Answer:

Position-Time graphs display the motion of a object by showing the changes of velocity with respect to time.

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2 years ago
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2 years ago
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A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.
Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is  1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

Learn more about energy here: brainly.com/question/13881533

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2 years ago
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4 0
3 years ago
How much momentum will a dumb-bell of mass 10 kg transfer
frosja888 [35]

We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:

Vf² = Vi² + 2ad

Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.

Given values:

Vi = 0m/s (dumbbell starts falling from rest)

a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)

d = 80×10⁻²m

Plug in the values and solve for Vf:

Vf² = 2(10)(80×10⁻²)

Vf = ±4m/s

Reject the negative root.

Vf = 4m/s

The momentum of the dumbbell is given by:

p = mv

p is its momentum, m is its mass, and v is its velocity.

Given values:

m = 10kg

v = 4m/s (from previous calculation)

Plug in the values and solve for p:

p = 10(4)

p = 40kg×m/s

6 0
3 years ago
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