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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.

Learn to know more about the risks of oversteering on

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A double-paned window has two glass panes separated by a layer of air. Why

EleoNora [17]

Answer:

for insulation

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3 years ago

Read 2 more answers

Read the following paragraph from the introduction

Alex17521 [72]

The statement 'energy cannot be created or destroyed' BEST supports the idea that energy remains constant during an energy transformation. It is the first law of thermodynamics.

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2 years ago

A playground merry-go-round of radius R = 1.20 m has a moment of inertia I = 240 kg · m2 and is rotating at 9.0 rev/min about a

Vesnalui [34]

Answer:

ω' = 0.815 rad/s

Explanation:

Given,

R = 1.20 m

Inertia of merry-go- round= 240 kg.m²

Rotating speed = 9 rpm = $9\times \dfrac{2\pi}{60}$

=0.9424 rad/s

mass of the child, m = 26 kg

angular speed of the merry-go-round=?

we know

Angular momentum, L = I ω

Moment of inertia of the child

I' = m r² = 26 x 1.2² = 37.44 kgm²

Conservation of angular momentum

initial angular momentum = Final angular momentum

I ω = (I+I')ω'

240 x 0.9424 = (240+37.44) ω'

226.176= 277.44 ω'

ω' = 0.815 rad/s

new angular speed of the merry-go- round is equal to 0.815 rad/s

3 0

4 years ago

While standing on an open bed of a truck moving at 35 m/s, an archer sees a duck flying directly overhead. The archer shoots an

pashok25 [27]

Answer:

Part a)

$t = 20 s$

Part b)

$x = 700 m$

Part c)

Since horizontal speed of truck and the arrow is same so the arrow will strike at the position of the archer as they both moving with same speed

Explanation:

Part a)

As we know that the arrow moves with uniform acceleration in vertical direction so we can use kinematics in Y direction

Here we know that

$\Delta y = v_y t + \frac{1}{2}at^2$

since the arrow lands at the same height so its vertical displacement of whole motion is zero

so we will have

$0 = 98 t - \frac{1}{2}(9.8) t^2$

$0 = 98 t - 4.9 t^2$

$t = 20 s$

Part b)

Horizontal distance moved by the arrow

$x = v_x t$

here horizontal speed of arrow is same as that of speed of truck

so we will have

$x = 35 \times 20$

$x = 700 m$

Part c)

Since horizontal speed of truck and the arrow is same so the arrow will strike at the position of the archer as they both moving with same speed

6 0

3 years ago

The tire of your bicycle needs air so you attach a bicycle pump to it and begin to push down on the pump’s handle. If you exert

dusya [7]

Answer:

12.5 J

Explanation:

Force, F = 25 N

Distance, d = 0.5 m

The direction of force and the displacement is same.

Work is defined as the product of force in the direction of displacement and the displacement.

Work = Force x displacement x CosФ

Where, Ф be the angle between force and the displacement

Here, Ф = 0°

So, W = 25 x 0.5 x Cos0°

W = 12.5 J

6 0

4 years ago