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miss Akunina [59]

4 years ago

15

_________ is affected by your brakes, tires, the road surface, and speed.

1 answer:

Rina8888 [55]4 years ago

6 0

Answer choices are:

A. Inertia

B. Friction

C. Kinetic Energy

D. Gravity

____________________________________________________________

Correct answer choice is:

B) Friction

____________________________________________________________

Explanation:

Friction is a force that takes back the action of a sliding object. Friction is just that simple. You will find resistance universally that things come into connection with each other. The force works in the opposing direction to the way an object wants to slide. It is the outcome of the electromagnetic attraction among charged particles in two moving surfaces.

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in what distance can a 1500kg automobile be stopped if the brake is applied when the speed is 20m/s and the coefficient of slidi

natali 33 [55]

1) The braking force is provided by the frictional force, which is given by:

$F_f=\mu m g$

where

$\mu=0.7$ is the coefficient of friction

m=1500 kg is the mass of the car

$g=9.81 m/s^2$ is the gravitational acceleration

Substituting numbers into the equation, we find

$F_f= (0.7)(1500 kg)(9.81 m/s^2)=10301 N$

2) The work done by the frictional force to stop the car is equal to the product between the force and the distance d:

$W=-F_fd$ (1)

where we put a negative sign because the force is in the opposite direction of the motion of the car.

3) For the work-energy theorem, the work done by the frictional force is equal to the variation of kinetic energy of the car:

$\Delta K=K_f -K_i =W$ (2)

The final kinetic energy is zero, so the variation of kinetic energy is just equal to the initial kinetic energy of the car:

$\Delta K=-K_i=-\frac{1}{2}mv^2=-\frac{1}{2}(1500 kg)(20 m/s)^2=-300000 J$

4) By equalizing eq. (1) and (2), we find the distance, d:

$-K_i = -Fd$

$d=\frac{K_i}{F}=\frac{300000 J}{10301 N}=29.1 m$

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3 years ago