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poizon [28]
1 year ago
5

Loss of traction between the rear wheels and road surfaces like ice, sand, or gravel results in?

Physics
1 answer:
kobusy [5.1K]1 year ago
6 0

Loss of traction between the rear wheels and road surfaces like ice, sand, or gravel results in oversteering.

Fishtailing is considered as a handling problem of a vehicle that occurs whenever traction is lost by the rear wheels which ends in oversteering. This can happen as a result of low friction surfaces including sand, gravel, rain, snow, and ice.

The traction loss of the rear tire can result in a state of oversteer. Whenever the rear tires do not have any grip on the surface of the road, steering a particular car will be having the effect of exaggerated results. Losing of rear wheel traction usually occurs at the time when the brake is applied while driving through a surface of a curved road.

In this case, a drop in the level of speed will be shifting weight into the front tires as it lessens the grip of the rear tire and which causes the rear end to swing out of the turn in opposite direction.

Oversteering is occurred by loss of traction between the rear wheels and road surfaces.

Learn to know more about the risks of oversteering on

brainly.com/question/6832517

#SPJ4

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Three cars (car F, car G, and car H) are moving with the same velocity when the driver suddenly slams on the brakes, locking the
Nastasia [14]

Answer:

Car H

Explanation:

Frictional force is a resistant force. It is given as:

F = u*m*g

Where u = coefficient of friction

m = mass

g = acceleration due to gravity

From the formula above, we see that frictional force is dependent on the mass of object and the coefficient of friction.

Since they all have the same tires, the coefficient of friction between the tire and the floor is the same for each car. Acceleration due to gravity, g, is constant.

The only factor that determines the frictional force of each car is the mass. Hence, the more the mass, the more the frictional force.

So, the most massive car will have the most frictional force and hence, will come to a stop quicker than the others. The least massive car will have the least frictional force and so, will take a longer time to stop.

5 0
2 years ago
horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
3 years ago
Read 2 more answers
Zoey ran a total of 1000m. 20 seconds into her run she had traveled 100 meters. What was her speed at 100 meters?
leva [86]
Let's assume that Zoey ran at a constant speed. we can use the equation,
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By rearranging,
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Hence, s = 100 m / 20 s = 5 m/s

therefore Zoey's speed at 100 m is 5 m/s
7 0
3 years ago
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murzikaleks [220]

Answer:

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Explanation:

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5 0
3 years ago
Read 2 more answers
A Ferris wheel has a diameter of 60 m and a period of rotation of 70 s. A passenger weighs 500 N. What is her apparent weight at
andrew11 [14]

Answer:

A

Explanation:

Hopefully this helps.

8 0
3 years ago
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