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1 year ago

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

1 answer:

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.

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3 years ago

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Marianna [84]

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3 years ago

Add these two velocity vectors to find the magnitude of their resultant vector.

hammer [34]

The magnitude of their resultant vector is 4.6 meters/seconds

Since we are to add the velocity vectors in order to find the magnitude of their resultant vector.

Hence:

Resultant vector magnitude=5.8 meters/seconds + (1.2 meters/seconds)

Resultant vector magnitude=5.8 meters/seconds-1.2 meters/seconds

Resultant vector magnitude 4.6 meters/seconds

Inconclusion The magnitude of their resultant vector is 4.6 meters/seconds

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3 years ago

Urgent!

Masja [62]

mass of iron block given as

$m_1 = 1.90 kg$

density of iron block is

$\rho = 7860 kg/m^3$

now the volume of the iron piece is given as

$V = \frac{m}{\rho}$

$V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3$

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

$F_b = \rho_L V g$

here we know that

$\rho_L$ = density of liquid = 916 kg/m^3

$F_b = 916* 2.42 * 10^{-4} * 9.8$

$F_b = 2.17 N$

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

$F_s + F_b = mg$

$F_s + 2.17 = 1.90* 9.8$

$F_s = 16.45 N$

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

$F_n = F_g + F_b$

$F_n = (1 + 2.50)*9.8 + 2.17$

$F_n = 34.3 + 2.17 = 36.47 N$

So the other scale will read 36.47 N

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3 years ago

Describe how freezing could be used to remove sugar from a mixture of sugar and water?

VARVARA [1.3K]

When the mixture (the sugar and water) is frozen, it separates. The water molecules get closer together, separating and pushing the sugar crystals to the top.<span />

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3 years ago

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