Asked by Troi
A car moves on a level horizontal road in
a circle of radius 30.5 m. the coefficient of
friction between the tires and the road is 0.5.
The maximum speed with which the car
can round this curve without slipping is:
a circle of radius 30.5 m. the coefficient of
friction between the tires and the road is 0.5.
The maximum speed with which the car
can round this curve without slipping is:
Answers
Answered by
Damon
F = m Acc
mu m g = m v^2/R
v^2 = mu g R
mu m g = m v^2/R
v^2 = mu g R
Answered by
Sophia
its 67
Answered by
Sinnamon
3 m/s
Answered by
simon
its clearly 13m/s
Answered by
Clara
v = sqrt( mu * g * r)
v = sqrt( .5 * 9.8 * 30.5)
v = 12.2m/s
v = sqrt( .5 * 9.8 * 30.5)
v = 12.2m/s
Answered by
Anonymous
The correct answer is 12.2 m/s
How?
To do it you need the equation μ x m x g = (m x v^2)/R
Since we are trying to find the velocity or the maximum speed of the car we would change the equation to v^2 = μ g R
μ = 0.5 which is the friction between the tires and the road
g = gravitational acceleration (9.8 m/s^2)
r = is the radius which is 30.5 m
Once plugged into the equation it should look like this:
v^2 = (0.5 x 9.8 x 30.5)
so v^2 = 149.45
you need to square root both sides so it would be v = √(149.45)
so the answer is 12.22497444 m/s but rounded it would be 12.22 m/s
How?
To do it you need the equation μ x m x g = (m x v^2)/R
Since we are trying to find the velocity or the maximum speed of the car we would change the equation to v^2 = μ g R
μ = 0.5 which is the friction between the tires and the road
g = gravitational acceleration (9.8 m/s^2)
r = is the radius which is 30.5 m
Once plugged into the equation it should look like this:
v^2 = (0.5 x 9.8 x 30.5)
so v^2 = 149.45
you need to square root both sides so it would be v = √(149.45)
so the answer is 12.22497444 m/s but rounded it would be 12.22 m/s
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