The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

$f_s=\frac{mv^{2}}{R}$

But we know that:

$f_s\leq \mu_s N$

And the normal force is given by the sum of the forces in the vertical direction:

$N-mg=0 \implies N=mg$

Finally, we have:

$f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27$

So, the minimum value for the coefficient of friction is 0.27.

4 0

3 years ago

You testify as an "expert witness" in a case involving an accident in which car A slid into the rear of car B, which was stopped

natulia [17]

F=mgsinα –F(fr)=

= mgsinα - μmgcosα

a=F/m =g(sinα – μcosα)

=9.8(sin14-0.57•cos14)= -3.04 m/s²

v²=v₀²+2ad

v=sqrt{ v₀²+2ad} =

=sqrt{18²+2• (-3.04) •25}=13.1 m/

3 0

2 years ago

What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-

sergey [27]

$q = -21 * 10^{-6} C$

<h3>What is Free-fall acceleration?</h3>

The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.

As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,

mg =qE

$1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\$

and its sign must be negative so that it will have upward electric force

so it is

$q = -21 * 10^{-6} C$

The charge of a particle of mass is $-21 * 10^{-6} C$

Learn more about Free-fall acceleration: brainly.com/question/2165771

#SPJ1

8 0

1 year ago