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What is the maximum speed with which a 1300 rubber-tired car can take this curve without sliding? (Take the static coefficient of friction of rubber on concrete to be 1.0.)

2007-11-04 07:39:38 · 4 answers · asked by venessadauby 1 in Science & Mathematics Physics

A concrete highway curve of radius 60.0m is banked at a 19.0 degree angle. What is the maximum speed with which a 1300kg rubber-tired car can take this curve without sliding? (Take the static coefficient of friction of rubber on concrete to be 1.0.)
The road is assumed to be dry.

2007-11-04 07:51:53 · update #1

4 answers

95mph

2007-11-04 07:41:31 · answer #1 · answered by RawrDinoAh 5 · 0 0

you cant prove that. the rubber is softer when when warm. and almost hard when cold. cant answer that questioin depends too much on the temp. of the the tires and the road.
19 degrees is pretty steep I say about 50 mph. I can do about 40 in my truck on a good day. but in wisconsin I did a 180 by acciddent. when the sign said slow to 20 I was going about 40 in a mustang. rubber gets pretty hard when its cold. it gets pretty sticky when hot. thats why they do "burn outs" before the drag races. to heat up the tires.

2007-11-04 07:51:51 · answer #2 · answered by hifi1863 2 · 0 1

The unfastened physique diagram shown on the listed reference depicts this properly. The centripetal tension vector necessary is: Fc = m*v^2/r This tension is point, in the x-course. The friction tension is: Ff = ?*N the place N is the conventional tension. The horizontal element of Ff is: Ffx = ?*N*Cos? The vertical element of Ff is: Ffy = -?*N*Sin? The horizontal element of the conventional tension is: Fnx = N*Sin? The vertical element of the conventional tension is: Fny = N*Cos? Write the equations for the x and y aspects, the x-aspects first: ?*N*Cos? + N*Sin? = m*v^2/r Now the y-aspects: N*Cos? - ?*N*Sin? = mg sparkling up the 1st equation for N: N * ( ?*Cos? + Sin? ) = m*v^2/r N = m*v^2/ [ r * ( ?*Cos? + Sin? ) ] replace into the 2nd equation: { m*v^2/ [ r * ( ?*Cos? + Sin? ) ] } * ( Cos? - ?*Sin?) = mg sparkling up for v: v = sqrt[ g*r * ( ?*Cos? + Sin? ) / ( Cos? - ?*Sin?) ] v = sqrt[ 9.8*60*(Cos(13) + Sin(13))/(Cos(13) - Sin(13)) ] = 30.sixty seven m/s it is examined using the calculator on the listed reference.

2016-11-10 06:30:40 · answer #3 · answered by piano 4 · 0 0

is the highway wet or dry

2007-11-04 07:42:32 · answer #4 · answered by the big chief 3 · 0 0

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