Ask question

Ask question

All categories

3 years ago

15

Rafael is driving his car at 26 m/s. What is the shortest distance in which he can brake and stop if the coefficient of static f

riction between the tires and the road is 0.4

1 answer:

Hitman42 [59]3 years ago

4 0

Answer:

86.14 meters.

Explanation:

Step one:

Given data

velocity of car = 26 m/s

the coefficient of static friction between the tires and the road

µ = 0.4 (kinetic)

Let us take g = 9.81 m/s^2

Required

The distance x = distance in m

We know that

$N = Fg = mg\\\\F_\mu = -\mu N = \mu F_g = \mu mg\\\\KE = (1/2) mv^2$

W = F*x (Work is force times distance)

Step two:

Conservation of energy gives

KE = W

Substituting gives

$(1/2) mv^2 = F \mu x\\\\(1/2) mv^2 = \mu mgx\\\\mv^2 = 2 \mu mgx$

Solving for distance (x) gives

$x = mv^2 / 2 \mu mg$

Simplifying

$x = v^2 / 2 \mu g$

Substitute:

$x = v^2 / 2 \mu g$

$x= 26^2/2*0.4*9.81$

$x=676/7.848\\\\x=86.14$

Therefore, the minimum braking distance is 86.14 meters.

You might be interested in

a 1.5 kg ball is thrown vertically upward with an initial speed of 15 m/s. if the initial potential energy is taken as zero, fin

trapecia [35]

Answer:

a) $E_{p} = 0$

$E_{k} = 168.7 J$

$E_{m} = 168.7 J$

b) $E_{p} = 73.6 J$

$E_{k} = 95.8 J$

$E_{m} = 169.4 J$

c) $E_{p} = 169.2 J$

$E_{k} = 0$

$E_{m} = 169.2 J$

Explanation:

We have:

m: is the ball's mass = 1.5 kg

v₀: is the initial speed = 15 m/s

g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

$E_{p} = mgh = 0$

The kinetic energy is:

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J$

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

$E_{p} = mgh = 1.5*9.81*5 = 73.6 J$

Now, to find the kinetic energy we need to calculate the speed at 5 m:

$v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9$

$v_{f} = \sqrt{126.9} = 11.3 m/s$

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(11.3)^{2} = 95.8 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J$

c) At its maximum height:

$v_{f}$ : is the final speed = 0

$h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m$

Now, the potential, kinetic and mechanical energies are:

$E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J$

$E_{k} = \frac{1}{2}mv^{2} = 0$

$E_{m} = 169.2 J + 0 = 169.2 J$

I hope it helps you!

7 0

2 years ago

An object of height 2.7 cm is placed 29 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 12

kati45 [8]

Answer:

A) q = -8.488 cm , B) m = 0.29

Explanation:

A) For this exercise in geometric optics, we will use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where p and q are the distance to the object and image, respectively and f is the focal length

in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

$\frac{1}{q} = \frac{1}{f} - \frac{1}{p}$

we calculate

$\frac{1}{q} = - \frac{1}{12} - \frac{1}{29}$

$\frac{1}{q}$ = - 0.1178

q = -8.488 cm

the negative sign indicates that the image is virtual

B) the magnification is given

$m = \frac{h'}{h} = - \frac{q}{p}$

we substitute

m = $- \frac{-8.488}{29}$

m = 0.29

the positive sign indicates that the image is right

4 0

3 years ago

The accepted standard measurement for mass is the A) liter. B) pound. C) Newton. D) kilogram.

andreev551 [17]

D) Kilograms is the standard unit of mass

6 0

3 years ago

agasfer [191]

Answer:

For a velocity versus time graph how do you know what the velocity is at a certain time?

Ans: By drawing a line parallel to the y axis (Velocity axis) and perpendicular to the co-ordinate of the Time on the x axis (Time Axis). The point on the slope of the graph where this line intersects, will be the desired velocity at the certain time.

_____________________________________________________

How do you know the acceleration at a certain time?

$Ans: We\ know\ that\ acceleration = \frac{Final\ velocity-Initial\ Velocity}{Time\ taken}$

Hence,

By dividing the difference of the Final and Initial Velocity by the Time Taken, we could find the acceleration.

_________________________________________________________

How do you know the Displacement at a certain time?

Ans: As Displacement equals to the area enclosed by the slope of the Velocity-Time Graph, By finding the area under the slope till the perpendicular at the desired time, we find the Displacement.

_________________________________________________________

4 0

2 years ago

Two blocks of clay, one of mass 1.00 kg and one of mass 7.00 kg, undergo a completely inelastic collision. Before the collision

RideAnS [48]

Answer:8 m/s

Explanation:

Given

$m_1=1 kg$

$m_2=7 kg$

kinetic Energy of $m_1=32 J$

initially $m_2$ is at rest and let say $m_1$ is moving with velocity u

kinetic Energy of $m_1 is =\frac{m_1u^2}{2}=32$

$u^2=64$

$u=8 m/s$

In Completely inelastic collision both mass stick together and move with common velocity

Suppose v is the common velocity

$m_1u+0=(m_1+m_2)v$

$v=\frac{1\times 8}{1+7}=1 m/s$

therefore Final velocity with which both blocks moves is 1 m/s

7 0

3 years ago