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igomit [66]
3 years ago
12

To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal

force and reduce vehicle traction it is important to

Physics
1 answer:
tatiyna3 years ago
5 0
Answer:  Add an incline or grade to the road track.

Explanation:
Refer to the figure shown below.

When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v =  linear (tangential) velocity to the circular path.

The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.

At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.

When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
 Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).


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kotegsom [21]

Answer:

b) se duplica

Explanation:

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8 0
3 years ago
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ale4655 [162]

Answer:

4) Increase

Explanation:

Hope this helps!

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6 0
2 years ago
A string is fixed at both ends and vibrating at 140 Hz, which is its third harmonic frequency. The linear density of the string
loris [4]

Answer:

Length of the string = 0.24 m

Explanation:

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L = (n/2f)√(T/μ)

L = (3/(2×140))√(2.3/0.0046) = 0.40 m

4 0
4 years ago
Read 2 more answers
Do only number 3 and thank
IrinaK [193]

Charge = 0.2 Ah is the correct answer...

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