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erik [133]
3 years ago
14

The mass of a sports car is 1000 kg. The shape of the car is such that the aerodynamic drag coefficient is 0.260 and the frontal

area is 2.10 m2. Neglecting all other sources of friction, calculate the initial acceleration of the car, if it has been traveling at 90 km/h and is now shifted into neutral and is allowed to coast. (Take the density of air to be 1.295 kg/m3.)
Physics
2 answers:
alisha [4.7K]3 years ago
8 0

Answer:

The initial acceleration is 0,221 m/s^2.

Explanation:

Fcar = Fairfriction\\m*a = 0.5*A*c*density*v^2\\a = 0.5*A*c*density*v^2 / 1000 kg = 0.5*2.10m^2*0.260*1.295kg/m^3*(25 m/s)^2 / 1000 kg = 0,221 m/s^2

Vinil7 [7]3 years ago
3 0

Answer:

-0.221m/s^2

Explanation:

We are given that

Mass of car=m=1000 kg

Drag coefficient,C_d=0.260

Frontal area,A=2.1 m^2

Speed, v=90 km/h=90\times \frac{5}{18}=25 m/s

1km/h=\frac{5}{18} m/s

Density of air,\rho=1.295 kg/m^3

We know that

Drag force,f_d=-\frac{1}{2}\rho v^2 AC_d

a=\frac{f_d}{m}

Using the formula

a=\frac{-\frac{1}{2}\times (1.295)\times (25)^2\times 2.1\times 0.26}{1000}=-0.221m/s^2

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Explanation:

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               1 / f = (n-1) (1 / R₁ - 1 / R₂)

                1 / f = (1.70 -1)) 1 / ∞ - 1/13)

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                 f = - 18.57 cm

Now we can use the constructor equation

             1 / f = 1 / o + 1 / i

             1 / i = 1 / f - 1 / o

              1 / i = -1 / 18.57 -1/20

               1 / i = -0.1038 cm

               I = -9.63 cm

For the height of the

image let's use magnification

                 m = h '/ h = - i / o

                  h ’= -h i / o

                  h ’= - 0.5 (-9.63) / 20

                  h ’= .0.24075 cm

b) we invert the lens

The focal length is

             1 / f = (1.70 -1) (1/13 - 1 / int)

              1 / f = 0.0538

             f = 18.57 cm

             1 / i = 1 / f -1 / o

             1 / I = 1 / 18.57 - 1/20

             1 / I = 3.85 10-3

             i = 259.74 cm

     

            h ’= - 0.5 259.74 / 20

             h ’= 6.4935 cm

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