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Rudik [331]
4 years ago
11

While traveling on a horizontal road at speed vi, a driver sees a large rabbit ahead and slams on the brakes. The wheels lock an

d the car begins to slide against the road. The car collides with the rabbit at a final speed of vf, after T seconds of braking. What is the coefficient of kinetic friction of the tires against the road?
Physics
1 answer:
MA_775_DIABLO [31]4 years ago
4 0

Answer:

μk = (Vf - Vc)/(T×g)

Explanation:

Given

Vi = initial velocity of the car

Vf = final velocity of the car

T = Time of application of brakes

g = acceleration due to gravity (known constant)

Let the mass of the car be Mc

Assuming only kinetic frictional force acts on the car as the driver applies the brakes,

The n from Newtown's second law of motion.

Fk = Mc×a

Fk = μk×Mc×g

a = (Vf - Vc)/T

Equating both preceding equation.

μk×Mc×g = Mc × (Vf - Vc)/T

Mc cancels out.

μk = (Vf - Vc)/(T×g)

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4 years ago
A car moving 20 m/s in the positive x-direction slams on the brakes and stops after skidding 20 m. what was its acceleration dur
pav-90 [236]
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v_f^2 - v_i^2 = 2 aS
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3 years ago
You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with
Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

           

Explanation:

For this exercise we use the constructor equation or Gaussian equation

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

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

Magnification a lens system is

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

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In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

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f = 6.0 cm

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

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

          q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

           h ’= h q / p

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           h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

 

f = 12 cm h ’= fo 12  cm

therefore we see that the lens with the shortest focal length has a smaller object

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4 years ago
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