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Tcecarenko
8 days ago
7

Which of the following statements are true for electric field lines? Check all that apply. Check all that apply. Electric field

lines point away from positive charges and toward negative charges. Electric field lines are continuous; they do not have a beginning or an ending. Electric field lines can never intersect. Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. At every point in space, the electric field vector at that point is tangent to the electric field line through that point.
Physics
1 answer:
Keith_Richards [3.2K]8 days ago
5 0

Answer:

Electric field lines radiate from positive charges and converge on negative charges. True

Electric field lines are continuous and do not have a start or end point. False

Electric field lines can never cross each other. True

In regions of space where the electric field is weak, the field lines are farther apart, and they are closer together in regions where it's strong. False

At any given point in space, the electric field vector aligns with the tangent to the electric field line at that location. True

Explanation:

Electric field lines emanate from positive charges and draw towards negative charges. They always move away from positive charges and head towards negative ones.

Electric field lines are continuous, without a definitive start or finish. False  

As the lines originate from positive charges and terminate at negative ones.

Electric field lines cannot cross one another. True

It’s not possible for them to intersect, as this would imply the field has multiple directions at one point. Moreover, the actual field value at that point would need to be determined by combining both field lines.

Electric field lines are spaced close together in areas where the electric field magnitude is potent and spaced apart in weak regions.False

This is incorrect as regions with higher field strength have lines that are closer together, while weaker regions have them farther apart.

At each location in space, the electric field vector coincides with the tangent to the corresponding field line at that point. True

This statement aligns with the definition of electric field lines.

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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
ValentinkaMS [3465]

Answer:

a) Blood mass is m= 5.7876kg

b) The count of blood cells is  N_t=1.04*10^{13}

Explanation:

From the problem statement, we learn that

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

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

         % of blood which consists of cells is  = 45.0%

        the % of blood that is  plasma is  = 55.0%

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

         % of cells that are white is  = 1%

        % of cells that are red is  = 99%

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

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

The mass is generally represented mathematically as

               m = \rho_b * V_b

Substituting values

            m = 1060 * 0.00546

               m= 5.7876kg

Cell mass 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 white blood cells volume is V_w = 1% of the cells volume

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

                                                       = 2.315*10^{-5}m^3

The volume of a single cell is V_s = 4 \pi r^3

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

                                                                        = 2.21*10^{-16}m^3

The red blood cells volume is V_r = V_c - V_w

                                                           =2.315*10^{-3} - 2.315*10^{-5}

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

The total red blood cell count is  = \frac{V_r}{V_s}

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

                                                    = 1.037*10^{13}

The total white blood cell count is   =\frac{V_w}{V_s}

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

                                                          = 1.04*10^{11}

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

                                                        N_t=1.04*10^{13}

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A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle
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To determine the average net force, we can calculate acceleration using:

x = 0.5*a*t^2

v = a*t

where x=3.6m and v=185 m/s.

Thus,

t=v/a and therefore x = 0.5*a*(v/a)^2 = 0.5 * (v^2)/a

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Calculate the average charge on arginine when ph=9.20. (hint : find the average charge for each ionizable group and sum these to
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