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Gekata [30.6K]
3 years ago
8

Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of poun

ds), and let y be the miles per gallon (mpg)
Engineering
1 answer:
Alex17521 [72]3 years ago
4 0

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.

Data Given:

x = 27 , 44 , 32 , 47, 23 , 40, 34, 52

y = 30, 19,  24,  13 , 29,  19,  21,  14

It is given that,

∑x = 299

∑y = 167

∑x^{2} = 11887

∑y^{2} = 3773

We are asked to verify the above values manually in this question.

So,

1. ∑x = 299

Let's verify it:

∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52

∑x = 299

Yes, it is equal to the given value. Hence, verified.

2. ∑y = 167

Let's verify it:

∑y = 30 + 19 +  24 + 13 + 29 + 19 +  21 +  14

∑y = 169

No, it is not equal to the given value.

3. ∑x^{2} = 11887

Let's verify it:

For this to find,  first we need to square all the value of x individually and then add them together to verify.

∑x^{2} = 27^{2} + 44^{2} + 32^{2} + 47^{2} + 23^{2} + 40^{2} + 34^{2} + 52^{2}

∑x^{2} = 11,887

Yes, it is equal to the given value. Hence, verified.

4. ∑y^{2} = 3773

Let's verify it:

Again, for this we need to find the squares of all the y values and then add them together to verify it.

∑y^{2} = 30^{2} + 19^{2} +  24^{2} + 13^{2} + 29^{2} + 19^{2} +  21^{2} +  14^{2}

∑y^{2}  = 3,845

No, it is not equal to the given value.

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mario62 [17]

Answer:

What is true about a point and shoot camera?

  • They are often smaller than SLR cameras

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3 years ago
Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes
loris [4]

Answer:

a). 8.67 x 10^{-3} m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L  = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X 10^{-5} Pa-s

We know density of air is ρ = 1.21 kg /m^{3}

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = \frac{\rho .U.x}{\mu }

Re = \frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = \frac{5.x}{\sqrt{Re}}

   = \frac{5\times 1}{\sqrt{332052.6}}

   = 8.67 x 10^{-3} m

a). Boundary layer thickness at x = 1 is δ = 8.67 X 10^{-3} m

b). Given Re = 100000

    Therefore the critical distance from the leading edge can be found by,

     Re = \frac{\rho .U.x}{\mu }

     100000 = \frac{1.21\times5\times x}{1.822 \times10^{-5}}

     x = 0.3011 m

c). Given x = 3 m from the leading edge

    The Reyonld no at x = 3 m from the leading edge

     Re = \frac{\rho .U.x}{\mu }

     Re = \frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }

     Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

    δ = \frac{0.38\times x}{Re^{\frac{1}{5}}}

       = \frac{0.38\times 3}{996158.06^{\frac{1}{5}}}

       = 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

  Re = \frac{\rho \times U\times x}{\mu }

5 X 10^{5} = \frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}

 x = 1.505 m

We know that the force acting on the plate is

F_{D} = \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

and C_{D} at x= 1.505 for a laminar flow is = \frac{1.328}{\sqrt{Re}}

                                                                         = \frac{1.328}{\sqrt{5\times10 ^{5}}}

                                                                       = 1.878 x 10^{-3}

Therefore, F_{D} =  \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

                                          = \frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}

                                         = 0.2137 N

e). The flow is turbulent at the end of the plate.

  Re = \frac{\rho \times U\times x}{\mu }

       = \frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }

       = 1992316

Therefore C_{D} = \frac{0.072}{Re^{\frac{1}{5}}}

                                           = \frac{0.072}{1992316^{\frac{1}{5}}}

                                           = 3.95 x 10^{-3}

Therefore F_{D} = \frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}

                                           = \frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}

                                          = 1.792 N

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(2) Asphalt : Used extensively to make Motor Road, highways

(3) Plastics : we use plastics in our everyday life, this is also a product of Refining of crude oil e.g PVC, Telephone casing, Tapes e.t.c

(4) Lubricating Oil/Grease : This is another product from crude oil Fractional Distillation.

(5) Propane/ Cooking Gas: This is also a product from oil which is used in our everyday life for cooking, grilling etc.

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