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Lelu [443]
3 years ago
7

How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 40.0°C greater th

an when they were laid? Their original length is 41.0 m.
Physics
1 answer:
ella [17]3 years ago
8 0

Answer:

19.68 × 10⁻³ m

Explanation:

Given;

Original Length, L₁ = 41.0 m

Temperature Change, ΔT = 40.0°C

Thermal Linear expansion of steel is given to be, ∝_{steel} = 12 × 10⁻⁶ /°C

   Generally, Linear expansivity is expressed as;

                               ∝ = ΔL / L₁ΔT

Where

∝ is the Linear expansivity

ΔL is the change in length, L₂ - L₁

L₂ is the final length

L₁ is the original length

ΔT is the change in temperature θ₂ - θ₁ (Final Temperature - Initial Temperature)              

From equation of linear expansivity

                         ΔL = ∝_{steel}L₁ΔT

                         ΔL = 12 × 10⁻⁶ /°C × 41.0 m × 40.0 °C

                         ΔL = 19.68 × 10⁻³ m

                         ΔL = 19.68 mm

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Answer:

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Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells
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Answer:

a

The mass of blood is m= 5.7876kg

b

The number of blood cells is  N_t=1.04*10^{13}

Explanation:

From the question we are told that

         The volume of blood  is  V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3

         The density of the blood is  \rho_b = 1060 kg/m^3

         % of blood  that is  cell is  = 45.0%

        % of the blood that is  plasma is  = 55.0%

        density of blood cell is  \rho_d = 1125kg/m^3

        % of cell that are white is  = 1%

        % of cell that is red is  = 99%

        The diameter of the red blood cell is  = 7.5 \mu m = 7.5*10^{-6}m

         The radius of the red blood cell is  = \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m

Generally the mass is mathematically  represented as

               m = \rho_b * V_b

Substituting value

            m = 1060 * 0.00546

               m= 5.7876kg

Mass of cell is m_c = 45% of m

                         = 0.45 * 5,7876

                         = 2.60442 kg

The volume of cells is V_c = \frac{m_c}{\rho_d}

                                      = \frac{2.60442}{1125}

                                      = 2.315 *10^{-3} m^3

The volume of white blood cell is V_w = 1% of volume of cells

                                                         = \frac{1}{100} * 2.315*10^{-3}

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                                                                        = 4*(3.142) * (3.75*10^{-6})^3

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The volume of red blood cells is V_r = V_c - V_w

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                                                           = 2.29*10^{-3}m^3

The number of red blood cell is  = \frac{V_r}{V_s}

                                                     = \frac{2.29 *10^{-3}}{2.21*10^{-16}}

                                                    = 1.037*10^{13}

The Number of white blood cell is   =\frac{V_w}{V_s}

                                                          = \frac{2.315 * 10^{-5}}{2.21*10^{-16}}

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The total number of blood cells is  N_t= 1.037*10^{13} + 1.04*10^{11}

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