Adriver in a 1000 kg car traveling at 20 m/s slams on the brakes and skids to a stop. if the coefficient of friction between the tires and the horizontal road is 0.80, how long will the skid marks be?

The car's skid marks will be 25.4842 m long.

Explanation:

According to Newton's second law:

Force = Mass × acceleration due to car

Also, The formula for frictional force,

Frictional force = μ × Normal Force

Also, Normal force = mass × acceleration due to gravitation(g)

So,

Frictional force = μ × mass × g

The two forces acting horizontally on the tire in opposite directions. So,

Mass × acceleration due to car = μ × mass × g

Solving,

Acceleration due to car = μ × mass × g

Given,

μ = 0.80

Also, 9.81 ms⁻²

So,

Acceleration due to car = 7.848 ms⁻²

Considering the Equation of motion as:

v² = u² - 2.a.s

Brakes are applied ad the car stops. The final velocity of the car (v) = 0 ms⁻¹

Given: Initial velocity of car (u) = 20 ms⁻¹

Acceleration, above calculated = 7.848 ms⁻²

Applying in the equation to calculate the distance as:

(0)² = (20)² - 2×(7.848)×s

So, Distance:

s = 25.4842 m

The skid marks are 25.4842 m long.

it becomes positively charged by charging it

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be