Turns along roads are banked at an angle so that vehicles can safely negotiate the turns. A turn with a radius of 300 m has a banking angle of 10°. What is the ideal velocity at which a vehicle can negotiate this turn?

Explanation:

For ideal velocity,

The centripetal force must be equal to the net horizontal force that balances the centripetal force

Given that,

Radius of path is r=300m

Banking angle = 10°

Centripetal acceleration is given as

Fc= mv²/r

And the net horizontal force is given as

Fnet, x=mgtanθ

Therefore,

Fc=Fnet, x

mv²/r = mgtanθ

Cross multiply

mv² = rmgtanθ

Divide from sides by m

v² = rgtanθ

Take square of both sides

v=√(rgtanθ)

Since r=300m, g=9.81m/s² & θ=10°

v=√(300×9.81×tan10)

v=√518.93

v=22.78m/s

the ideal velocity at which a vehicle can negotiate this turn is 22.78m/s

where are the vectors ?

as we probably am aware that our atmosphere or air have various levels, each level has different gasses, and each gas has another porousness. at the point when the light originates from lower porousness to higher penetrability, it alters its course. with the goal that implies that when the light originates from the moon to the earth, it alters it's course a lot more occasions. it essentially implies that the state of the light piece of the moon is changed on the earth.

not copied too! hope this !

joke:

h o h o hope you have a merry christmas!