The frictional force required is 9000 N
Explanation:
In order to keep the car in the turn in circular motion without sliding, the frictional force must provide the centripetal force necessary for the circular motion.
Therefore, we can write:

where the term on the left is the frictional force while the term on the right is the centripetal force, and where:
m is the mass of the car
v is its speed
r is the radius of the curve
For the car in this turn, we have
m = 1000 kg
v = 30 m/s
(since the diameter is 0.20 km, the radius is half that value)
And substituting, we find

Learn more about friction:
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The weights in newtowns for the given masses are
<span> masses 22.1, 33.5, 41.3, 59.2, 78
weights 216.58N 328.3N 404.74N 580.16N 764.4N
e.g, for m=22.1kg, W=22.1kgx9.8N/kg =216.58N</span>
Answer:
47.5kJ
Explanation:
Before climbing the cliff


At 2.
