Physics, 29.07.2019 06:00 hiccup5653oye1pb

Acar is traveling at 52.0 mi/h on a horizontal highway. (a) if the coefficient of static friction between road and tires on a rainy day is 0.102, what is the minimum distance in which the car will stop? m (b) what is the stopping distance when the surface is dry and µs = 0.603? m

Answers: 1

Show answers

Answers

Answer from: dondre54

First of all, let's convert the initial speed from mi/h to m/s.
We know that 1 mi=1609 m and 1 h=3600 s. So, the conversion factor is:
$1 \frac{mi}{h}= 1 \frac{1609 m}{3600 s}=0.45$
And the initial speed of the car is therefore:
$v_i = 52 mi/h \cdot 0.45 = 23.4 m/s$
Now we can start to solve the problem.

a) I assume that 0.102 is the coefficient of dynamic friction, not static friction, because the car is in motion.
The only force acting on the car is the frictional force, directed against the motion of the car. We can write Newton's second law for the car:
$ma=-\mu mg$
where m is the mass of the car, a its acceleration, and $\mu m g$ is the frictional force, with a negative sign since it is acting against the motion of the car. Simplifying m and solving, we can find the acceleration of the car:
$a=-\mu g=-(0.102)(9.81 m/s^2)=-1 m/s^2$
The acceleration is negative because the car is decelerating.

Now we can use the following relationship valid for a uniformly accelerated motion:
2aS = v_f^2-v_i^2

where S is the distance covered, vi and vf are the initial and final speed of the car. Since the car is going to stop, v_f=0

, and by replacing a and viwith the values we calculated previously, we can find the distance S that the car needs to stop:
$S=- \frac{(v_i)^2}{2a}= -\frac{(23.4 m/s)^2}{2(-1 m/s^2)}=273.8 m$

b) The procedure to solve this part of the exercise is exactly identical to the first part, the only difference is the value of the coefficient of friction: $\mu=0.603$ . Let's start by calculating the new acceleration
$a=-\mu g=-(0.603)(9.81 m/s^2)=-5.9 m/s^2$
And then, the new distance needed to stop the car is
$S=- \frac{(v_i)^2}{2a}=- \frac{(23.4 m/s)^2}{2(-5.9 m/s^2)}=46.4 m$
So, when the surface is dry the car needs much less space to stop completely than when the surface is wet.

Answer from: Quest

option number c is the right answer.

Answer from: Quest

it's when something is going so fast it makes a loud noise, breaking the sound barrier, for example: a jet makes a loud noise, not only cause of the jet but because they reached a point where they went so fast that you can year the noise from various parts, kinda like a police car, lets say the car was going so fast that you hear the siren of the cop car right next to you, but the car is probably a mile a way, it's cos the siren was next to you a few seconds if you understand me

Another question on Physics

Physics, 22.06.2019 04:30

Which of the following is not a characteristic of s waves? a. travel slower than p waves. b. cannot be detected in locations more than 105° from an earthquake’s epicenter. c. travel through solids and liquids. d. only affect coastal regions.

Answers: 2

Answer

Physics, 23.06.2019 00:30

3. which is not a primary color of light? blue red green yellow

Answers: 1

Answer

Physics, 23.06.2019 02:40

The two metal spheres in the figure are connected by a metal wire with w switch in the middle. initially the switch is open. sphere 1, with the larger radius, i'd given a positive charge. sphere 2, with the smaller radius, is neutral. then the switch is closed. afterward, sphere 1 has charge q1, is at potential v1 and the electric field strength at its surface is e1. the values for sphere 2 are q2, v2, and e2. a. is v1 < ,> , or = to v2? explain. b. is q2 < , > , or = explain? c. is e1 < ,> ,or = to e2? explain?

Answers: 1

Answer

Physics, 23.06.2019 05:00

How long will it take a car traveling of a constant speed of 45m/s to cover the distance of a 2.5km?

Answers: 1

Answer

You know the right answer?

Acar is traveling at 52.0 mi/h on a horizontal highway. (a) if the coefficient of static friction be...

Questions