How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius 90 m at a speed of 90 km/h ?

Explanation:

Given data

the radius of curve= 90m

speed v= 90 km/h to m/s = (90*100)/60*60=  25 m/s

we know that the expression for the centripetal force acting on the car

-------1

we also know that the expression for the frictional force between road and tire.

Ff= μmg--------2

Equating equation 1 and 2 we have

μmg= mv^2/r

μ= v^2/gr

substituting the values of speed and radius we have (assuming g= 9.81m/s^2)

μ= 25^2/9.81*90

μ= 625/882.9

μ= 0.7

if the thermos has a vacuum innner, then heat through conv currents and conduction are reduced = vacuum flask

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j