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2 years ago

6

Immediately after it has started to rain, dust particles and oil residue float on the water and ____________ the traction of tir

es.

1 answer:

Mkey [24]2 years ago

8 0

When it rains, dust particles and oil residues float on the water and this reduces the traction of tires.

<h3>What is traction?</h3>

This concept refers to a force between the tires and road that causes the movement of the wheels or vehicle is slower.

<h3>What happens with traction when it rains?</h3>

It is well-known more accidents occur when it rains, which is caused by cars slidding on the road. This is because when it rains traction or the grip of the wheel drastically reduces.

Learn more about traction in: brainly.com/question/14525337

You might be interested in

A rocket is fired at 100 m/s at an angle of 37, how many seconds did it take to get to the top?

Kobotan [32]

Answer:

6.14 s

Explanation:

The time the rocket takes to reach the top is only determined from its vertical motion.

The initial vertical velocity of the rocket is:

$u_y = u sin \theta = (100)(sin 37^{\circ})=60.2 m/s$

where

u = 100 m/s is the initial speed

$\theta=37^{\circ}$ is the angle of launch

Now we can apply the suvat equation for an object in free-fall:

$v_y = u_y +gt$

where

$v_y$ is the vertical velocity at time t

$g=-9.8 m/s^2$ is the acceleration of gravity

The rocket reaches the top when

$v_y =0$

So by substituting into the equation, we find the time t at which this happens:

$t=-\frac{u_y}{g}=-\frac{60.2}{-9.8}=6.14 s$

7 0

3 years ago

A 0.80 kg object tied to the end of a 2.0 m string swings as a pendulum. At the lowest point of its swing, the object has a kine

devlian [24]

Answer:

3.32 m/s

Explanation:

From the law of conservation of energy, the sum of mechanical and kinetic energy should be equal to the 10 J given. Potential energy is given by mgh where m is mass, g is acceleration due to gravity and h is the height. For this case, $h= l(1-cos\theta)$ and l is string length, given as 2 m, \theta is given as 50 degrees. Kinetic energy is given by $0.5mv^{2}$ and it is this velocity that is unknown.

$10J=0.5\times 0.8kg\times v^{2}+ 0.8kg\times 9.81\times 2m(1-cos 50^{\circ})\\v\approx 3.32 m/s$

6 0

3 years ago

A 40.0 -kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 13

Irina18 [472]

The work done by the gravitational force = 0

Given the mass of the box = 40 kg

The box is initially at rest.

Distance moved by the applied force = 5m

Force applied = 130 N

Co-efficient of friction between the box and floor = 0.3

The box is moved only in the horizontal direction by the applied force.

Gravitational force is applied in a direction perpendicular to the applied force. hence it doesn't do any work on the box.

Therefore, the work done by the gravitational force is zero.

Learn more about the gravitational force at brainly.com/question/862529

#SPJ4

3 0

2 years ago

While playing her guitar , karen plucks one string with increasing levels of force. What effect does this have on the sound prod

Lena [83]

Answer:

The amplitude of vibration of string will increase due to which loudness of sound will increase

Explanation:

As we know that the guitar is based on the principle of Resonance. When string of the guitar vibrates at a given frequency then the sound produced in the hollow part of the guitar will also be at same frequency.

This is known as resonance condition, so guitar will produce same frequency sound as that of frequency of string.

Now if the string is plucked with increasing level of force then it will increase the amplitude of vibrations of the string due to which the sound produced in the guitar will also be of same level.

So here we can say that amplitude and intensity of sound related as

$I = kA^2$

so on increasing amplitude the intensity will increase and hence it will produce loud sound

8 0

3 years ago

Read 2 more answers

A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subj

KATRIN_1 [288]

Answer:

$\Delta t = 8 s$

Explanation:

As we know that the angular acceleration of the wheel due to friction is constant

so we can use kinematics

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

so we have

$(65 \times 2\pi) = (2\pi \times 9)(10) + \frac{1}{2}(\alpha)(10^2)$

$130\pi = 180\pi + 50 \alpha$

$\alpha = -\pi rad/s^2$

now time required to completely stop the wheel is given as

$\omega_f = \omega_i + \alpha t$

$0 = (2\pi \times 9) + (-\pi) t$

$t = 18 s$

now time required to stop the wheel is given as

$\Delta t = 18 - 10$

$\Delta t = 8 s$

6 0

3 years ago