Asked by aleen
meerany
A flat (unbanked) curve on a highway has a radius of 220.0
Ill. A car rounds the curve at a speed of 25.0 m/s. (a) What is the minimum
coefficient of friction that will prevent sliding? (b) Suppose the highway is
icy and the coefficient of friction between the tires and pavement is only
one-third what you found in part (a). What should be the maximum speed of
the car so it can round the curve safely?
can some one help me in this question..
A flat (unbanked) curve on a highway has a radius of 220.0
Ill. A car rounds the curve at a speed of 25.0 m/s. (a) What is the minimum
coefficient of friction that will prevent sliding? (b) Suppose the highway is
icy and the coefficient of friction between the tires and pavement is only
one-third what you found in part (a). What should be the maximum speed of
the car so it can round the curve safely?
can some one help me in this question..
Answers
Answered by
MathMate
Centripetal force
=mv²/r
friction force
=μmg
Equate and solve for μ
μ=mv²/r / mg
=v²/gr
For the second part, reduce μ to one-third, and try the reverse calculation.
=mv²/r
friction force
=μmg
Equate and solve for μ
μ=mv²/r / mg
=v²/gr
For the second part, reduce μ to one-third, and try the reverse calculation.
Answered by
nk
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Answered by
imnotsayingmyname
i just had to do this for my one class, so i’ll try to explain it.
using the equation, substitute the numbers from the equation in.
so you should have 25 / 220 * 9.8.
multiply 220 by 9.8 to get 2,156. then divide 25 by 2,156 to get 0.01159.
ANSWER: 0.011
using the equation, substitute the numbers from the equation in.
so you should have 25 / 220 * 9.8.
multiply 220 by 9.8 to get 2,156. then divide 25 by 2,156 to get 0.01159.
ANSWER: 0.011
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