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3 years ago

How does speed affect the friction between a road and a skidding tire?

Physics

1 answer:

Step2247 [10]3 years ago

3 0

<span>speed does not affect friction, friction affects force</span>

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an object of mass m is rotating about a fixed axis with angular momentum l. its moment of inertia about this axis is i. what is

Tems11 [23]

The Kinetic energy would be 1/2IL².

<h3>What is Rotational Kinetic energy ?</h3>

Rotational energy also known as angular kinetic energy is defined as: The kinetic energy due to the rotation of an object and is part of its total kinetic energy. Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.

As we know linear Kinetic energy = 1/2mv²

where m= mass and v= velocity.

Similarly rotational kinetic energy is given by = 1/2IL²

where I- moment of inertia and L=angular momentum.

To know more about the Kinetic energy , visit:

brainly.com/question/29807121

#SPJ4

8 0

1 year ago

Emily uses a rifle to shoot a bullet at a target. The bullet has a mass of 13 grams. The rifle has a mass of 3,500 grams. When s

sp2606 [1]

Force exerted by the bullet = mass * acceleration = 0.013 * 850 = 11.05 Newtons.

the rifle exerts same force in opposite direction so we have

11.05 = 3.5 * a
acceleration = 11.05 / 3.5 = 3.16 m /s^-2

4 0

3 years ago

Read 2 more answers

A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting

ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

$T=\dfrac{38}{8}\\\\T=4.75\ s$

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

k = spring stiffness constant

So,

$k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m$

When the cord is in air,

mg=kx

x = the extension in the cord

$x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m$

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

5 0

3 years ago

Which of the following best describes the location of the

marshall27 [118]

Answer:

The mantle exists above the crust of the earth

8 0

3 years ago

A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

$E_{initial} = E_{final}$

$PE_{initial}+KE_{initial} = PE_{final}+KE_{final}$

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

$KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}$

$KE_{final} = (5100+4200)-5700$

$KE_{final} = 3600MJ$

Therefore the final kinetic energy is 3600MJ

5 0

4 years ago