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r-ruslan [8.4K]
1 year ago
9

Chen is testing the friction of three surfaces. he pushes the same ball across three different surfaces with the same force and

measures the distance the ball rolls over each surfacel. the ball moved ten inches across surface 1, six inches across surface 2, and fifteen inches across surface 3. which could most likely describe the three surfaces? surface 1 is ice, surface 2 is gravel, and surface 3 is blacktop. surface 1 is gravel, surface 2 is ice, and surface 3 is blacktop. surface 1 is blacktop, surface 2 is gravel, and surface 3 is ice. surface 1 is blacktop, surface 2 is ice, and surface 3 is gravel.
Physics
1 answer:
e-lub [12.9K]1 year ago
5 0

<u>Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.</u> is most likely describe the three surfaces.

<h3>What is friction?</h3>

Friction, force that resists the sliding or rolling of one solid object over another. Frictional forces, such as the traction needed to walk without slipping, may be beneficial, but they also present a great measure of opposition to motion. About 20 percent of the engine power of automobiles is consumed in overcoming frictional forces in the moving parts.

The major cause of friction between metals appears to be the forces of attraction, known as adhesion, between the contact regions of the surfaces, which are always microscopically irregular. Friction arises from shearing these “welded” junctions and from the action of the irregularities of the harder surface plowing across the softer surface.

Learn more about friction

brainly.com/question/1424758

#SPj4

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude
Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

\omega_f = \omega_i + \alpha t

where

\omega_i = 0.300 rev/s is the initial angular velocity

\alpha = 0.895 rev/s^2 is the angular acceleration

Substituting t = 0.200 s, we find

\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s

Let's now convert it into rad/s:

\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

r=\frac{0.800}{2}=0.400 m

And so now we can find the tangential speed at t = 0.200 s:

v=\omega_f r =(3.01)(0.400)=1.2 m/s

2) 2.25 m/s^2

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

a_t = \alpha r

where \alpha is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

a_t = (5.62)(0.400)=2.25 m/s^2

3) 3.6 m/s^2

The centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

a=\frac{1.2^2}{0.400}=3.6 m/s^2

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LARGER vibration produces a higher ............
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In the year 1178. five monks at Canterbury Cathedral in England observed what appeared to be an asteroid colliding with the moon
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Answer: 1.28 sec

Explanation:

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