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2 years ago

What braking technique slows the vehicle as quickly as possible without locking brakes or losing traction

1 answer:

8 0

Answer: Cadence braking or Stutter braking is the braking technique slows the vehicle as quickly as possible without locking brakes or losing traction.

Explanation: To find the answer, we need to know more about the Cadence breaking.

<h3>What is cadence braking?</h3>

It's a braking technique, that includes, pumping the brake pedal and which used to allow a vehicle to steer as well as brake on a slippery surface.
It is used in the case of an emergency or sudden braking is needed.
In this technique, traction is limited to reduce the effect of skidding from road wheels locking up under braking.

Thus, from the above given data, we can conclude that, Cadence braking or Stutter braking is the braking technique slows the vehicle as quickly as possible without locking brakes or losing traction.

Learn more about the cadence braking here:

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Two small plastic spheres are given positive electrical charges. When they are 16.0 cm apart, the repulsive force between them h

Drupady [299]

Answer:

q = 7.542 x 10⁻⁷ C = 754.2 nC

Explanation:

The Coulomb's Law gives the magnitude of the force of attraction or repulsion between two charges:

F = kq₁q₂/r²

where,

F = Force of attraction or repulsion = 0.2 N

k = Coulomb's Constant = 9 x 10⁹ N m²/C²

r = distance between charges = 16 cm = 0.16 m

q₁ = magnitude of 1st charge

q₂ = magnitude of 2nd charge

Since, both charges are said to be equal here.

q₁ = q₂ = q

Therefore,

0.2 N = (9 x 10⁹ N m²/C²)q²/(0.16 m)²

(0.2 N)(0.16 m)²/(9 x 10⁹ N m²/C²) = q²

q = √(5.88 x 10⁻¹³ C²)

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3 years ago

Inside a NASA test vehicle, a 3.50-kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The forc

erastova [34]

F(of spring)=230x=ma=3.5(5)=17.5=230x; x=0.07m.

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3 years ago

A 70 kg object is on a table in static equilibrium. What is the support (normal) force the table exerts on the box

sweet-ann [11.9K]

F(n) = mg = 70 x 9.8 = 68.6 N

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4 years ago

Canyon walls are eroded at a rate of approximately:

AnnyKZ [126]

B. Half a centimeter per year

God bless!! <3

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3 years ago

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

$F=mrw^2$

$m=$ mass of the corresponding bugs

$r=$ corresponding radial distance of each bug

$w=$ angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

$F$ ∝ $mr$

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

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3 years ago

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