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Gnom [1K]
3 years ago
15

A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 N. What is the Young's modul

us of this wire?
A) 6.43 × 109 N/m2
B) 1.41 × 1010 N/m2
C) 5.66 × 1010 N/m2
D) 2.78 × 109 N/m2
E) 3.54 × 109 N/m2
Physics
1 answer:
n200080 [17]3 years ago
3 0

Answer:

(B) 1.41 x 10^{10} N/m^{2}

Explanation:

diameter of wire (d) = 0.6 mm = 0.0006 m

radius of wire (r) = 0.0006/2 = 0.0003 m

tension (force F) = 20 N

percentage strain = 0.5 %

what is the Young's modulus?

Young's modulus = \frac{stress}{strain}

where

  • stress = force/area

        stress = \frac{force}{πr^{2}}

        stress = \frac{20}{πx0.0003^{2}}

        stress =70735530.3 N/m^{2}

  • strain = percentage strain ÷ 100

        strain = 0.5 ÷ 100 = 0.005

therefore Young's modulus = \frac{stress}{strain} =\frac{70735530.3}{0.005} = 1.41 x 10^{10} N/m^{2}

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Tanya [424]

Answer:

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4 0
2 years ago
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
barxatty [35]

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}

                                                       = 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 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}}

                                                    = 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}}

                                                          = 1.04*10^{11}

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

                                                        N_t=1.04*10^{13}

6 0
4 years ago
Read 2 more answers
Ryan runs 50 m north and then 50 m east. What is his displacement?
san4es73 [151]

Answer:

Explanation:

Distance and direction of an object's change in position from a starting point. Displacement. 3. Jermaine runs exactly 2 laps around a 400 meter track. What is the displacement? 0 ... David walks 3 km north, and then turns east and walks 4 km. ... A person walks 50 meters directly north, stops, and then travels 32 meters

3 0
3 years ago
What are 3 ways acceleration can change?
Karolina [17]

Velocity, direction, or both at the same time.

3 0
3 years ago
A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a
AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

q = -Kb\frac{dh}{dl}

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25  m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

        = 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours

4 0
3 years ago
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