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

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop