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beks73 [17]
3 years ago
7

Define the types of friction and give FOUR examples of each

Physics
1 answer:
Verdich [7]3 years ago
4 0

Answer:

Static Friction - acts on objects when they are resting on a surface

Sliding Friction -  friction that acts on objects when they are sliding over a surface

Rolling Friction - friction that acts on objects when they are rolling over a surface

Fluid Friction - friction that acts on objects that are moving through a fluid

Explanation:

Examples of static include papers on a tabletop, towel hanging on a rack, bookmark in a book , car parked on a hill.

Example of sliding include sledding, pushing an object across a surface, rubbing one's hands together, a car sliding on ice.

Examples of rolling include truck tires, ball bearings, bike wheels, and car tires.

Examples of fluid include water pushing against a swimmer's body as they move through it , the movement of your coffee as you stir it with a spoon,  sucking water through a straw, submarine moving through water.

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In deep space, there is very little friction. Once they launch a probe into deep space, where there are no external forces actin
Charra [1.4K]

Answer:

move at constant velocity.

Explanation:

Newton's first law (also known as law of inertia) states that:

"when the net force acting on an object is zero, the object will keep its state of rest or if it is moving, it will continue moving at constant velocity".

In the case of the probe, friction in deep space is negligible, therefore when the engine is shut down, there are no more forces acting on the probe: the net force therefore will be zero, so the probe will move at constant velocity.

5 0
3 years ago
Read 2 more answers
Chord progressions that move to resting points that release tension are called
Sav [38]

Cadences.

These cadences are the resulting tensions that chords release from their resting points. This movement is classified from a unstable chord progression to a stable one.  Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
6 0
2 years ago
The joule (J) is a unit of energy. Recall that energy may be converted between many different forms such as mechanical energy, t
REY [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is  W = -177.275J

Explanation:

From the question we are told that

      The initial Volume is  Vi = 0.160 L

      The final volume is  V_f = 0.510L

      The external pressure is  P = 5.00 \ atm

Generally the change in volume is

           \Delta V = V_f - V_i

Substituting values we have

           \Delta V = 0.510 -0.160

                 = 0.350L

Generally workdone is mathematically represented as

           W = -P \Delta V

W is negative because the working is done on the environment by the system which is indicated by volume increase

     Substituting values

                W = - 5* 0.350

                    = -1.75 \ L \ \cdot atm

Now  1 \  L \cdot atm = 101.3J

  Therefore  W = -1.75* 101.3

                          = -177.275J

   

7 0
2 years ago
Read 2 more answers
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
-Dominant- [34]

Explanation:

LD₁ = 10⁵ mm⁻²

LD₂ = 10⁴mm⁻²

V = 1000 mm³

Distance = (LD)(V)

Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m

Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m

Conversion to miles:

Distance₁ = 10×10⁴ m / 1609m = 62 miles

Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.

7 0
3 years ago
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determine the loudness (in decibels) of the sound at a rock concert if the intensity of the sound is 1 x 10–1 w/m2. remember, th
EleoNora [17]

The loudness of the sound at the rock concert, where the intensity of the sound is1 x 10⁻¹ Wm⁻² is  110 dB.

Here we are dealing with loudness which is the perception of the Intensity of the sound.

The formula  to refer to in order  to  find the value of the loudness of a sound is ,

  db= 10log(I/I₀)

As we are provided with the current intensity which is  1 x 10⁻¹ Wm⁻². and the initial intensity which is  1 x 10⁻¹² Wm⁻².

So, by substituting the required values in the formula we get

db= 10 * log( 1 x 10⁻¹ /1 x 10⁻¹²)

 = 10 * 11 log(10)

 = 110

So, the result is 110 dB.

To know more about the intensity of sound refer to the link brainly.com/question/9323731?referrer=searchResults.

To know more about questions related to loudness refer to the link brainly.com/question/21094511?referrer=searchResults.

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4 0
1 year ago
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