From the given options the best example of an ascribed status will be poor. Hence, option 2 is the correct one.
What is ascribed status?In sociology, the term "ascribed status" refers to a human's social standing as it is either allocated to them at birth or implicitly assumed in later life. The position of status is one that a person neither chooses for themselves nor earns. Instead, the position is decided upon in accordance with cultural and societal norms and standards. No matter how hard you try or how much you want to, these slots are filled. These fixed social designators are independent of the favorable or unfavorable stereotypes associated with one's assigned statuses and stay unchanged throughout a person's life.
Every society engages in the process of classifying people into categories based on factors including gender, race, family history, and ethnicity. This process is cross-cultural.
To know more about ascribed status:
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Rank the following circuits in order from highest to lowest values of the current in the circuit.
i. a 1.4-Ω resistor connected to a 1.5-V battery that has an internal resistance of 0.10 Ω;
ii. a 1.8-Ω resistor connected to a 4.0-V battery that has a terminal voltage of 3.6 V but an unknown internal resistance;
iii. an unknown resistor connected to a 12.0-V battery that has an internal resistance of 0.20 Ω and a terminal voltage of 11.0 V.
a. (i), (iii), (ii)
b. (iii), (i), (ii)
c. (ii), (iii), (i)
d. (i), (ii), (iii)
e. (iii), (ii), (i)
f. (ii), (i), (iii)
Answer:
e)
Explanation:
Let's get first the values of the currents for the three cases.i)
The battery forms a series circuit with its internal resistance and the 1.4 Ω resistor. Since the current is the same at any point of the circuit, and the sum of all voltages along a closed circuit must be zero, we can apply Ohm's Law in each resistor, as follows:[tex]V = I*r_{i} + I*R_{1} (1)[/tex]
Replacing V, ri and R₁ by their values, we can solve for the current I as follows:[tex]I_{i} = \frac{V}{r_{int} + R_{i}} = \frac{1.5V}{0.1 \Omega + 1.4 \Omega} = 1.0 A (2)[/tex]
ii)
Since the voltage of the battery is 4.0 V (open circuit voltage), and it falls to 3.6 V when is connected to a 1.8Ω resistor, this means that the voltage through the resistor must be 3.6 V, due to the sum of all voltages along a closed circuit must be zero.So, we can find the current through the circuit, applying Ohm's Law to the 1.8Ω resistor, as follows:[tex]I_{ii} =\frac{V_{term} }{R_{ii} } =\frac{3.6V}{1.8 \Omega} = 2.0 A (3)[/tex]
iii)
Since the 12.0 V battery has a terminal voltage of 11.0 , this means that the voltage through the internal resistance of 0.2 Ω, must be 1.0 V.So we can find the current Iiii, applying Ohm's Law to the internal resistance value, as follows:[tex]I_{iii} =\frac{V-V_{term}}{r_{int} } =\frac{12.0 V- 11.0 V}{0.2 \Omega} =\frac{1.0V}{0.2\Omega} = 5.0 A (4)[/tex]
So, the highest current is the Iiii, followed by Iii and Ii, which is stated by e).Which phrase best completes the diagram?
Goals of Social Policy:
Providing citizens with an education ➡️Ensuring the safety of citizens➡️?
A. Helping people who live in poverty.
B. Increasing aid to foreign countries.
The correct answer is:
A. Helping people Who live in poverty
Answer:
Did you ask the question and answer it for yourself?
Answer:
Your amazing thank you, A was right.
The force of friction depends upon
Answer:
ask internet?
Explanation:
easy .....
..
A sound having a frequency of 299 Hz travels through air at 332 m/s.
What is the wavelength of the sound? Answer in units of m.
Answer:
1.11 m.
Explanation:
Why?
The speed for a wave is done by the equation: v = f * w. Because the frecuency tells us about how many cycles the wave makes each time, but for each cycle the wave runs certain distance, given for the wavelenght. If you isolate the letter w you get the value just doing a ratio.
v = speed
f = frecuency
w = wavelenght
w = v / f
what is the magnitude of electric field at apoint 2 meter from a point charge q= 4nc
Alternate current
Hello!
I need solve 2 questions, I tried but im so bad at physics :(
1. If the frequency in the network is 50Hz, what is the period of this voltage?
2. The effective voltage value in the network is 230V. What is its maximum value?
I would like an explanation with a calculation for my understanding!!
Thanks in advance you all!
sorry bro but i don't know his i am also bad
Explanation:
but maybe you can you the formula of volatage as well as
The National Institute of Health is testing sensors which measure the energy felt by goalkeepers when blocking soccer balls coming at them and the velocity of soccer balls. One sensor measuring velocity is inside a 16 ounce soccer ball, which is kicked by a player at an initial velocity of 31 m/s. The ball bounces off the goalkeeper at a final velocity of 2.25 m/s. How much energy in Joules should the sensor say the goalkeeper absorbed, not accounting for wind and drag
Answer:
E = 216.76 J
Explanation:
The amount of energy absorbed by the goalkeeper will be equal to the difference between the initial and the final kinetic energy of the ball.
[tex]Energy\ Absorbed = Initial\ Kinetic\ Energy - Final\ Kinetic\ Energy\\Energy\ Absorbed = E = \frac{1}{2}mv_{i}^2 - \frac{1}{2}mv_{f}^2\\E = \frac{1}{2}m(v_{i}^2 - v_{f}^2)\\[/tex]
where,
m = mass of ball = (16 ounce)(0.0283495 kg/1 ounce) = 0.4535 kg
vf = final velocity = 2.25 m/s
vi = initial velocity = 31 m/s
Therefore,
[tex]E = \frac{1}{2}(0.4535\ kg)[(31\ m/s)^2 - (2.25\ m/s)^2][/tex]
E = 216.76 J
Help please thank you
Outline 3 disadvantages and advantage of water and alcohol as a template liquid
Answer:
Advantages of mercury as a thermometric liquid.
-It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.
-It does not wet (cling to the sides of) the tube.
-It has a high boiling point
-It expands uniformly (linear expansion) and responds quickly to temperature changes, hence is sensitive.
-It has a visible meniscus.
Disadvantages
-Mercury is very poisonous.
-its expansively is fairly low
-it is expensive
-It has a high freezing point therefore it cannot be used in places where the temperature gets very low.
Alcohol has a thermometric fluid
-Alcohol expands uniformly.
-It has a low freezing point (-115 degreecentigrade) therefore it is very suitable for place where the temperature gets very low.
-It has a large expansively
-It is an easily available cheap liquid, which is safe to use
Disadvantages of alcohol
-it wets the tube
-it has a low boiling point (cannot be used in places with high temperatures)
-it does not react quickly to changes in temperature
-It needs to be dyed, since it's colourless.
Disadvantages of water
-Water has high specific heat capacity. So it cannot be used for measuring small temperature differences.
- Water will wet the surface of the glass tube. It is a sticky substance.
- Water is transparent
Explanation:
Match the following items.
1. Extremely small building blocks of matter.
2. Forming new matter from old matter.
3. Small bits of matter.
atom
molecule
chemical change
An aspirin tablet that contains 75mg of
aspirin and 325mg of inert materials is an
example of
A. qualitative data.
B. quantitative data.
C. neither qualitative or quantitative data.
Answer:
it is for sure B. Quantitative data !
Explanation:
As I learned from quizlet!. there your welcome
Answer:
B. Quantity
Explanation:
This is easy
An important news announcement is transmitted by radio waves to people who are 46 km away, sitting next to their radios, and by sound waves to people sitting across the newsroom, 2.1 m from the newscaster. Take the speed of sound in air to be 348 m/s. What is the difference in time that the message is received
Answer:
[tex]0.005847\ \text{s}[/tex]
Explanation:
Radio waves travel at the speed of light
[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
[tex]d_r[/tex] = Distance between two radios = 46 km
[tex]v[/tex] = Speed of sound in air = 348 m/s
[tex]d_s[/tex] = Distance sound travels across the newsroom = 2.1 m
Time taken by the radio signal to reach the required location is
[tex]t_r=\dfrac{d_r}{c}\\\Rightarrow t_r=\dfrac{46\times 10^3}{3\times 10^8}\\\Rightarrow t_r=0.000153\ \text{s}[/tex]
Time taken by sound to reach the required location is
[tex]t_s=\dfrac{d_s}{v}\\\Rightarrow t_s=\dfrac{2.1}{348}\\\Rightarrow t_s=0.006\ \text{s}[/tex]
The time difference is
[tex]t_s-t_r=0.006-0.000153=0.005847\ \text{s}[/tex]
The difference in time that the message is received is [tex]0.005847\ \text{s}[/tex].
. Four railroad cars, each of mass 2.50 104 kg, are coupled together and coasting along horizontal tracks at a speed of vi toward the south. A very strong but foolish movie actor, riding on the second car, uncouples the front car and gives it a big push, increasing its speed to 4.00 m/s southward. The remaining three cars continue moving toward the south, now at 2.00 m/s. (a) Find the initial speed of the cars. (b) How much work did the actor do
Answer:
a) v₀ = 2.5 m / s, heading south.
b) W = 1,219 10⁵ J
Explanation:
a) For this exercise we can use the conservation of the moment, we create a system formed by all the 4 cars, in this case when the last one separates the forces are intense and the moment is conserved
initial instant. Before separation
p₀ = M v₀
final instant. When uncoupling the last car
p_f = 3m v₁ + m v₂
where they indicate that the speed of the wagons is v₁ = 2.00 m / s and the speed of the last wagon is v₂ = 4.00 m / s
the total mass is M = 4m
how the moment is preserved
p₀ = p_f
4m v₀ = 3m v₁ + mv₂
v₀ = ¾ v₁ + v₂ / 4
let's calculate
v₀ = ¾ 2 + ¼ 4
v₀ = 2.5 m / s
heading south.
b) work is equal to the change in kinetic energy
W = ΔK = K_f -K_o
W = ½ m v_f² - ½ m v₀²
W = ½ m (v_f² -v₀²)
W = ½ 2.50 10⁴ (4² - 2.5²2)
W = 1,219 10⁵ J
The provided text reads: "a typical lightning bolt may transfer 10^20 electrons in a fraction of a second, developing a peak current of up to 10 kiloamperes." Using the value of the elementary charge of 1.6x10^-19 C We can estimate the total charge of the lightning bolt to be about *
1.6E1 C
1.6E4 C
1.6E10 C
1.6E20 C
Answer:
1.6e20
Explanation:
Assume a uniformly charged ring of radius R and charge Q produces an electric field Ering at a point P on its axis, at distance x away from the center of the ring. Now the charge Q is spread uniformly over the circular area the ring encloses, forming a flat disk of charge with the same radius. How does the field Edisk produced by the disk at P compare to the field produced by the ring at the same point?Assume a uniformly charged ring of radius R and charge Q produces an electric field Ering at a point P on its axis, at distance x away from the center of the ring. Now the charge Q is spread uniformly over the circular area the ring encloses, forming a flat disk of charge with the same radius. How does the field Edisk produced by the disk at P compare to the field produced by the ring at the same point?
Answer:
* E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]
*E_ disk= 2kQ [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]
Explanation:
Let's start by finding the electric field of the charged ring
in the attachment we can see a diagram of the system. Due to circular symmetry, the electric field perpendicular to the axis is canceled and only the electric field remains parallel to the axis.
Eₓ = E cos θ (1)
E = k ∫ [tex]\frac{dq}{r^2}[/tex]
cos θ = x / r
using the Pythagorean theorem
r = [tex]\sqrt{x^2 + y^2}[/tex]
we substitute
Eₓ = k ∫ [tex]\frac{dq}{x^2+y^2} \ \frac{x}{\sqrt{ x^2+y^2} }[/tex]
Eₓ = [tex]k \frac{x}{(c^2+y^2)^{3/2} }[/tex] ∫ dq
Eₓ = k \frac{x}{(c^2+y^2)^{3/2} } Q
the ring's electric field
E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]
Now let's find the electric field of the disk
The charge is distributed over the entire disk, so let's use the concept of charge density
σ = [tex]\frac{dq}{dA}[/tex]
Let's approximate the disk as a group of rings, the width of each ring is dr, the area is
dA = 2πr dr
we substitute
σ = [tex]\frac{1}{2\pi r} \ \frac{dq}{dr}[/tex]
dq = 2π σ r dr
we substitute in equation 1, where the electrioc field is of each ring
Eₓ = [tex]k \int\limits^R_0 \ { \frac{x}{(x^2+r^2)^{3/2} } \ 2\pi \sigma \ r } \, dr[/tex]
if we use a change of variable
dv = 2rdr
v = r²
Eₓ = [tex]k x \pi \sigma \int\limits^a_b { \frac{1}{(x^2+v)^{3/2} } } \, dv[/tex]
we integrate
Eₓ = k x π σ [tex][ \frac{ (x^2 + r^2)^{-1/2} }{-1/2} ][/tex]
we value in the limits from r = 0 to r = R
Eₓ = k π σ x (-2) [ [tex]\frac{1}{ \sqrt{x^2+R^2} } - \frac{1}{x}[/tex]]
Eₓ = 2π k σ ([tex]1 - \frac{x}{(x^2 + R^2 ) ^{1/2} }[/tex] )
σ = Q/πR²
substitute
Eₓ = 2 k Q/R² (1 - \frac{x}{(x^2 + R^2 ) ^{1/2} } )
E_ disk= 2kQ [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]
The two electric fields are
* E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]
*E_ disk= 2kQ [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]
we can see that the functional relationship of the two fields is different
In fatal crashes, more than __________% of passenger car occupants who were totally ejected from the vehicle were killed.
Answer:
it would be 83% in a fatal crash.
Which of these is NOT a form of energy?
A. mechanical
B. chemical
C. theoretical
D. thermal
Answer:
I think it's theoretical.
10. A 50 kg bicyclist on a 10 kg bicycle speeds up from 5.0 m/s to 10 m/s.
(a) What was the total kinetic energy before accelerating? Full working out
Answer:
T.K.E = 750 Joules.
Explanation:
Given the following data;
Initial velocity, u = 5m/s
Final velocity, v = 10m/s
Mass of bicyclist = 50kg
Mass of bicycle = 10kg
Total mass, Tm = 50 + 10 = 60kg
Kinetic energy can be defined as an energy possessed by an object or body due to its motion.
Mathematically, kinetic energy is given by the formula;
[tex] K.E = \frac{1}{2}mv^{2}[/tex]
Where;
K.E represents kinetic energy measured in Joules.
m represents mass measured in kilograms.
v represents velocity measured in metres per seconds square.
To find the total kinetic energy before accelerating simply means the kinetic energy due to the initial velocity and total mass;
[tex] T.K.E = \frac{1}{2}T_{m}U^{2}[/tex]
Substituting into the equation, we have;
[tex] T.K.E = \frac{1}{2}*60*5^{2}[/tex]
[tex] T.K.E = 30*25 [/tex]
T.K.E = 750 Joules.
Arrange the following substances from the lightest to the heaviest:
Cl2, CH4; H20; NH3, N2
A. H2O<NH,< N2 CH4Cl2
B. CH< NH< H-0< < Cl2
C. Ny< Cl< H2O< CH«<NH3
D. NH;< CH«< Cl< H2O< N2
molecular weights are written in the picture.
CH4<NH3<H2O<Cl2
Marcy pulls a backpack on a wheels down the 100m hall. The 60N force is applied at an angle of 30° above the horizontal. How much work is done by Marcy?
Answer:
Work= Fcos∆×S
W=60N×cos 30⁰×100
W=60×0.866×100=5196.1J
PLEASE GIVE BRAINLIEST
27. The electric field around a positive charge is shown in the diagram. Describe the nature of these lines. Please use 2 content related sentences
Answer:
Once we place a positive test at a point close to the sphere, we find that an electrostatic force is applied to the outside of the sphere. Therefore, at any point around the sphere, the electric field vector is radially outward.
5. It takes
to get a job done.
Answer:
determination
Explanation:
g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first bounce and do NOT assume the total energy is the same as the total energy of the object before it hits the ground.) a. List the quantities that you need to know to determine the total energy of the object after it hits the ground. b. Record your measurements and describe how you measured them. c. Calculate the total energy of the object after it hit the ground. Your final answer: ______________ d. Determine whether or not the object’s energy was conserved when it hit the ground. If it was not conserved, explain where the energy went.
Answer:
a) quantity to be measured is the height to which the body rises
b) weighing the body , rule or fixed tape measure
c) Em₁ = m g h
d) deformation of the body or it is transformed into heat during the crash
Explanation:
In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.
a) What quantities must you know to calculate the energy after the bounce?
The quantity to be measured is the height to which the body rises, we assume negligible air resistance.
So let's use the conservation of energy
starting point. Soil
Em₀ = K = ½ m v²
final point. Higher
Em_f = U = mg h
Em₀ = Em_f
Em₀ = m g h₀
b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.
c) We use conservation of energy
starting point. Soil
Em₁ = K = ½ m v²
final point. Higher
Em_f = U = mg h
Em₁ = Em_f
Em₁ = m g h
d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.
There are two possibilities.
* that have been equal therefore energy is conserved
* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash
What is the magnitude of the resultant vector? Round
your answer to the nearest tenth.
m
R
5 m
13 m
Intro
Done
Answer: 13.9 m
Explanation:
Answer:
13.9m
Explanation:
Answer on Edge
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.10 s for the boat to travel from its highest point to its lowest, a total distance of 0.700 m. The fisherman sees that the wave crests are spaced a horizontal distance of 6.10 m apart. A. How fast are the waves traveling?B. What is the amplitude of each wave? C. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, how fast are the waves traveling? D. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, what is the amplitude of each wave?
Answer:
a) v = 2.9 m / s, b) A = 0.350 m, c) v = 2.9 m / s, d) A = 1.00 m
Explanation:
The oscillatory motion is described by the expression
x = A cos (wt + Ф)
the wavelength which is the distance for the wave to repeat and the frequency which is the number of times a wave oscillates per unit of time
a) In this part they ask us for the speed of the wave.
Let's use the relationship between speed, wavelength and frequency
v = λ f
For the wavelength they indicate that the distance between two crest is 6.1 m
λ / 2 = 6.10
λ = 12.20 m
They give us the period of the wave is the time it takes to return to the same point, in this case they give half a period
A / 2 = 2.10
A = 4.20 me
f = 1 / t
f = ¼, 2
f = 0.238 Hz
let's calculate
v = 12.20 0.238
v = 2.9 m / s
b) the amplitude of the wave, is the distance from zero to some maximum
2A = 0.700
A = 0.350 m
c) the speed of the wave is not function of the amplitude, so the speed is the same
v = 2.9 m / s
d) the amplitude is
2A = 0.50
A = 1.00 m
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal different information. You will use the both momentum principle and the energy principle in this problem.
A satellite of mass 3500 kg orbits the Earth in a circular orbit of radius of 7.3 106 m (this is above the Earth's atmosphere).The mass of the Earth is 6.0 1024 kg.
What is the magnitude of the gravitational force on the satellite due to the earth?
F= ________N
Using the momentum principle, find the speed of the satellite in orbit. (HINT: Think about the components of (dp^^\->)\/(dt) parallel and perpendicular to p^^\->.)
v = ________ m/s
Using the energy principle, find the minimum amount of work needed to move the satellite from this orbit to a location very far away from the Earth. (You can think of this energy as being supplied by work due to something outside of the system of the Earth and the satellite.)
work= ________J
Answer:
A) F_g = 26284.48 N
B) v = 7404.18 m/s
C) E = 19.19 × 10^(10) J
Explanation:
We are given;
Mass of satellite; m = 3500 kg
Mass of the earth; M = 6 x 10²⁴ Kg
Earth circular orbit radius; R = 7.3 x 10⁶ m
A) Formula for the gravitational force is;
F_g = GmM/r²
Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²
Plugging in the relevant values, we have;
F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²
F_g = 26284.48 N
B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.
Thus;
GmM/r² = mv²/r
Making v th subject, we have;
v = √(GM/r)
Plugging in the relevant values;
v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))
v = 7404.18 m/s
C) From the energy principle, the minimum amount of work is given by;
E = (GmM/r) - ½mv²
Plugging in the relevant values;
E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)
E = 19.19 × 10^(10) J
If an engine cannot be 100%
efficient, what happens to the
energy that is lost?
A. It disappears.
B. It reproduces.
C. It still exists, but in a different form.
Answer: Given the evidence in the explanation, I'm pretty sure it's C. It still exists, but in a different form.
Explanation: "Some part of the energy supplied is used to change the internal energy of the system. Some part is also released into the surroundings. Generally, frictional losses are more predominant for the machines being not 100% efficient. This friction leads to the loss of energy in the form of heat, into the surroundings."
A diagram of a plant cell is shown below.
Which organelle is found in both plant and animal cells?
Cell membrane
Chloroplast
Ο Ο Ο
Large Vacuole
Cell Wall
Answer:
The Cell Membrane is found in both plant and animal cells.
Explanation:
Chloroplast, Large Vacuole and Cell Wall are all found in plant cells.
27. The traffic officer issued violation tickets to traffic
violators. If 68 of them composed 80% of the total
violators, how many violators are there in all?
A. 83
B. 84
C. 85
D. 86
28. There are 18 roses in a bunch of 24 flowers.
What percent of the flowers are roses?
A. 18%
B. 24% C. 60% D. 75%
Please help me ❤️❤️ thank u
Hope god
Answer:
27: 85
28:75%
Explanation:
27:68=80
?=100 hence (68×100)÷80
=85
28:18/24× 100
=75%
Your heart pumps blood into your aorta (diameter 2.5 cm) with a maximum flow rate of about 500 cm^3/s. Assume that blood flow in the aorta is laminar (which is not a very accurate assumption) and that blood is a Newtonian fluid with a viscosity similar to that of water.
a. Find the pressure drop per unit length along the aorta. Compare the pressure drop along a 10 cm length of aorta to atmospheric pressure (105 Pa).
b. Estimate the power required for the heart to push blood along a 10 cm length of aorta, and compare to the basal metabolic rate of 100 W.
c. Determine and sketch the velocity profile across the aorta (assuming laminar flow). What is the velocity at the center
Answer:
a. i) The pressure drop per unit length is 52,151.89 Pa
ii) The atmospheric pressure ≈ 19.175 × The pressure drop along 10 cm length of aorta
b i) The power required for the heart to push blood along a 10 cm length of aorta, is 2.6075945 Watts
ii) The basal metabolic rate ≈ 38.35 × The power to push the blood along a 10 cm length of aorta
c. i) Please find attached the drawing for the velocity profile created with Microsoft Excel
ii) The velocity at the center is approximately 2.04 m/s
Explanation:
The given diameter of the aorta, D = 2.5 cm = 0.025 m
The maximum flow rate, Q = 500 cm³/s = 0.0005 m³/s
Assumptions;
The blood flow is laminar
The blood is a Newtonian fluid
The viscosity of water ≈ 0.01 poise = 1 cp
a. i) The pressure drop per unit length of pipe ΔP/L is given by the Hagen Poiseuille equation as follows;
[tex]Q = \dfrac{\pi \cdot R^4}{8 \cdot \mu} \cdot \left(\dfrac{\Delta p}{L} \right)[/tex]
Where;
Q = The flow rate = A·v
A = The cross sectional area
R = The radius = D/2
Δp/L = The pressure drop per unit length of the pipe
Therefore, we have;
[tex]\dfrac{\Delta p}{L} = \dfrac{Q\cdot 8 \cdot \mu }{\pi \cdot R^4} = \dfrac{0.0005 \times 8 \times 1}{\pi \times 0.0125^4 } = 52151.89[/tex]
The pressure drop per unit length ΔP/L = 52,151.89 Pa
ii) The pressure, ΔP, drop along 10 cm (0.1 m) length of aorta = ΔP/L × x;
∴ ΔP = 52,151.89 Pa × 0.1 m = 5,215.189 Pa
Given that the atmospheric pressure, [tex]P_{atm}[/tex] = 10⁵ Pa, we have;
[tex]P_{atm}[/tex]/ΔP = 10⁵/5,215.189 ≈ 19.175
Therefore, the atmospheric pressure is approximately 19.175 times the pressure drop along 10 cm length of aorta
b. i) The power, P = Q × ΔP
Therefore, the power required for the heart to push blood along a 10 cm length of aorta, is P₁₀ = 0.0005 m³/s × 5,215.189 Pa = 2.6075945 Watts
ii) Therefore compared to the basal metabolic rate of, 'P', 100 W, we have;
P/P₁₀ = 100 W/2.6075945 Watts = 38.349521 ≈ 38.35
The basal metabolic rate is approximately 38.35 times more powerful than the power to push the blood along a 10 cm length
c. i) The velocity profile across the aorta is given as follows;
[tex]v_m = \dfrac{1}{4 \cdot \mu} \cdot \dfrac{\Delta P}{L} \cdot R^2[/tex]
Where;
[tex]v_m[/tex] = The velocity at the center
We get;
[tex]v_m = \dfrac{1}{4 \times 1} \times 52,151.89 \times 0.0125^2 \approx 2 .04[/tex]
The velocity at the center, [tex]v_m[/tex] ≈ 2.04 m/s
ii) The velocity profile, v(r), is given by the following formula;
[tex]v(r) = v_m \cdot \left[1 - \dfrac{r^2}{R^2} \right][/tex]
Therefore, we have;
[tex]v(r) = 2.04 - \dfrac{2.04 \cdot r^2}{0.0125^2} \right] = 2.04 - 163\cdot r^2[/tex]
The velocity profile of the pipe is created with Microsoft Excel