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Bond [772]
3 years ago
12

A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it

from behind with a force of 190 N, continually pushing with this force through a distance of 8 m. The astronaut moves around to the front of the crate and slows the crate down by pushing backwards with a force of 170 N, backing up through a distance of 6 m. After these two maneuvers, what is the speed of the crate?
Physics
1 answer:
Darina [25.2K]3 years ago
7 0

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given:  </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.  

<u>Solution  </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next  

Ef=Ei+W                                                 (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).  

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form  

(Kf + mc^2) = (KJ+ mc^2)                       (2)  

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.  

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W                              (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by  

W = FΔr                                                      (4)  

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut  

W = F Δr= (190N)(6 m) = 1140 J  

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate  

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,  

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W'                                 (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by  

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate  

1/2mvf'^2=1/2mvi'^2+W'

  vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

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Answer:

187.38 m

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Using the equation of motion

s = ut + 1/2gt²...................... Equation 1

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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

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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}

                                                       = 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

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

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

                                                           = 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}}

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

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The total 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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