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SashulF [63]
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
Physics
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.

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Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
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Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

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7 0
2 years ago
As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

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We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

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Finally the answer is 6.3 N at 162° counterclockwise from  

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