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m v 2 r = μ s mg . size 12{m { {v rSup { size 8{2} } } over {r} } =μ rSub { size 8{s} } ital "mg"} {}

We solve this for μ s size 12{μ rSub { size 8{s} } } {} , noting that mass cancels, and obtain

μ s = v 2 rg . size 12{μ rSub { size 8{s} } = { {v rSup { size 8{2} } } over { ital "rg"} } } {}

Solution for (b)

Substituting the knowns,

μ s = ( 25.0 m/s ) 2 ( 500 m ) ( 9 . 80 m/s 2 ) = 0 . 13 . size 12{μ rSub { size 8{s} } = { { \( "25" "." 0" m/s" \) rSup { size 8{2} } } over { \( "500"" m" \) \( 9 "." "80 m/s" rSup { size 8{2} } \) } } =0 "." "13"} {}

(Because coefficients of friction are approximate, the answer is given to only two digits.)

Discussion

We could also solve part (a) using the first expression in F c = m v 2 r F c = mr ω 2 } , size 12{ left none matrix { F rSub { size 8{c} } =m { {v rSup { size 8{2} } } over {r} } {} ##F rSub { size 8{c} } = ital "mr"ω rSup { size 8{2} } } right rbrace ,} {} because m , size 12{m,} {} v , size 12{v,} {} and r size 12{r} {} are given. The coefficient of friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is a responsive force, being able to assume a value less than but no more than μ s N size 12{μ rSub { size 8{g} } N} {} . A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below.

In the given figure, a car is shown from the back, which is turning to the left. The weight, w, of the car is shown with a down arrow and N with an up arrow at the back of the car. At the right rear wheel, centripetal force is shown along with its equation formula in a leftward horizontal arrow. The free-body diagram shows three vectors, one upward, depicting N, one downward, depicting w, and one leftward, depicting centripetal force.
This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

Let us now consider banked curves , where the slope of the road helps you negotiate the curve. See [link] . The greater the angle θ size 12{θ} {} , the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an “ideally banked curve,” the angle θ size 12{θ} {} is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression for θ size 12{θ} {} for an ideally banked curve and consider an example related to it.

For ideal banking    , the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions.

[link] shows a free body diagram for a car on a frictionless banked curve. If the angle θ size 12{θ} {} is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight w size 12{w} {} and the normal force of the road N size 12{N} {} . (A frictionless surface can only exert a force perpendicular to the surface—that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude mv 2 /r size 12{"mv" rSup { size 8{2} } "/r"} {} . Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force—that is,

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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Abhi
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kinnecy Reply
can someone help me with some logarithmic and exponential equations.
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ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
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is it a question of log
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salma
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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what is nano technology
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what is system testing?
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preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Azam
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Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
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not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
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I'm interested in Nanotube
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this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Physics subject knowledge enhancement course (ske). OpenStax CNX. Jan 09, 2015 Download for free at http://legacy.cnx.org/content/col11505/1.10
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