Answer:
Did you ask the question and answer it for yourself?
Answer:
Your amazing thank you, A was right.
describe how electrons create electricity
describe how electrons create electricity
Atoms are made up of even smaller elements, called protons, electrons and neutrons. When electrical and magnetic forces move electrons from one atom to another, an electrical current is formed
Electrons in atoms can act as our charge carrier, because every electron carries a negative charge. If we can free an electron from an atom and force it to move, we can create electricity. ... In its balanced state, copper has 29 protons in its nucleus and an equal number of electrons orbiting around it.
Hope It Helps!
[1 mark]
29. A ball rolls at a constant speed of 3.0 m/s for 8.0 seconds. How far does it roll in this time!
A. 2.7 m
B. 3.0 m
C. 8.0 m
D. 24.0 m
Please help me
Answer:
D. 24.0m
Explanation:
formula
S=d/t
s=speed
d=distance
t=time
s=3.0m/s
d=x
t=8.0s
s=d/t
d=s*t
d=3.0*8.0
d=24.0m
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
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
the moon's mass is ____?
Answer:
i hope this is helpful
Explanation:
7.34767309 × 1022 kilograms
Answer:
Hellooo
7.35 x 1022 kg
The moon's mass is 7.35 x 1022 kg, about 1.2 percent of Earth's mass. Put another way, Earth weighs 81 times more than the moon.
Explanation:
thanks
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.
You hold your physics textbook in your hand. (Assume that no other objects are in contact with the book.)
a. Identify the forces acting on the book.
1. book on hand
2. hand on book
3. floor on book
4. earth on book
b. For each force you identified in part (a), indicate the direction. (If a force is not applicable, choose NA.)
1, book on hand __________
2. hand on book __________
3. floor on book __________
4. earth on book _________
Answer:
a) Please see below as the answer is self-explanatory.
b) 1. vertically downward
2. vertically upward
3 N/A
4 vertically downward.
Explanation:
a)
If the book is hold by the hand, this means (according to Newton's third law) that both exert a force on each other, equal and opposite.In this case this pair is formed by normal forces, the normal force (upward) that the hand does on the book (avoiding it to be accelerated downward) and the normal force that the book does on the hand (downward) avoiding it to be accelerated in the upward direction.The other force that acts on the book is the one due to gravity, and is always downward, which is numerically equal to the normal force that the hand does on the book, due to the book is at rest. There is no interaction between the book and the floor, so there is no force between them, neglecting the gravitational attraction between their masses.An 80 N force causes a spring to compress 0.15 m. What is the spring constant? What is the potential energy of the spring?
The spring constant of the given spring is= -533.33
The potential energy of the spring= 6 Joule
What is force?
A push or pull that causes object to change its state of rest or motion.
It can also change the direction of a moving body.Force is a product of mass and acceleration.For spring;
K is the spring constant.
⇒Force = -kx
⇒80N=0.15m ×-k
⇒K=-80/0.15
⇒K= -533.333
So the spring constant is 533.333
Potential Energy=1/2kx²
⇒0.5×533.33 ×0.15×0.15
⇒5.95.99 N
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ANSWER DIS ASAP FOR LUCK TMRW :}
List down ten situations that show how friction affects the movement of objects. I NEED THE ANSWER IN ONE MINUTE
Answer:
Walking –We can walk only if we apply frictional force. Friction is what holds your shoe to the ground. The friction present on the ice is very little, this is the reason why it is hard to walk on the slippery surface of the ice.
Writing – A frictional force is created when the tip of the pen comes in contact with the surface of the paper. Rolling friction is what comes into play while writing with a ballpoint pen while sliding friction arises when one writes with a pencil.
Skating – A thin film of water under the blade is necessary to make the skate slide. The heat generated by the skate blade rubbing against the surface of ice causes some of the ice to melt right below the blade where the skater glides over the ice. This water acts as a lubricant reducing friction.
Lighting a matchstick – When the head of the matchstick is rubbed against a rough surface, heat is generated and this heat converts red phosphorous to white phosphorous. White phosphorous is highly inflammable and the match stick ignites. Sometimes, matchsticks fail to ignite due to the presence of water. Water lowers friction.
Driving of the vehicle on a surface- While driving a vehicle, the engine generates a force on the driving wheels. This force initiates the vehicle to move forwards. Friction is the force that opposes the tyre rubber from sliding on the road surface. This friction avoids skidding of vehicles.
Applications of breaks in the vehicle to stop it- Friction braking is the most widely used braking method in vehicles. This process involves the conversion of kinetic energy to thermal energy by applying friction to the moving parts of a vehicle. The friction force resists the motion and in turn, generates heat. This conversion of energy eventually bringing the velocity to zero.
Flight of aeroplanes- Drag is the force that opposes the forward motion of the aeroplane. The friction which resists the motion of an object moving through a fluid or immobile in a moving fluid, as occurs when we fly a kite. The friction of the air is created as it meets and passes over an aeroplane and its components. Drag is generated by air impact force, skin friction and displacement of the air.
Drilling a nail into the wall- Friction is responsible for fixing of nails in a wall. As the nail is driven into the wall, the nearby material to the nail of the wall gets compressed. This exerts a force on the nail. This force is the friction that converts the normal force exerted by the compressed layers of the wall into the resisting shear force. In this manner the friction cause nails and screws to hold on to walls.
The dusting of the carpet by beating it with a stick- When the carpet is beaten with the stick, the dust comes out. The dust is carried off by the wind or falls on the floor. The carpet exhibits a little static friction that holds the dust to the carpet. When the carpet is beaten, it will overcome the friction and the carpet will move away from the dust making the carpet free from dust.
Sliding on a garden slide- We know that friction is a force that is present whenever two objects rub against each other. In case of a slide in the garden such as a slide and a person’s backside rub each other’s surface. Without friction, a slide would accelerate the rider too quickly, resulting in possible injury due to the fall. The friction reduces the velocity of the sliding person and makes him stop.
Hope that helps! :D Sorry if it's too lengthy...
Explanation:
write the type of energy present in pond water and kerosene?
The energy present in pond water and kerosene are potential energy and chemical energy respectively.
What is energy?Energy is a quantifiable quantity in physics that can be transferred from an item to do work. Thus, we might characterise energy as the capacity to engage in any kind of physical action. So, to put it simply, we can define energy as: Energy is the capacity to carry out work.
Energy "can neither be created nor destroyed but can only be changed from one form to another," according to the rules of conservation of energy. The SI unit for energy is called a joule.
In pond, water is stored during rain and stored potential energy whereas in kerosene, carbon molecules have chemical energy and when it burns, chemical energy revels as thermal energy.
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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
Which letter on the map represents the
Southern Ocean
Answer:
B cus is in the south but that pfp tho
. 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
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
PLEASE HELP ME WITH THIS ONE QUESTION
1000 J of energy is given to a balloon via a heating device. Assuming the pressure is a constant 101,325 Pa, what is the final volume of the balloon, assuming it started at 0.001 m3?
Answer:
Vf = 0.0108 m³
Explanation:
Assuming there is no change in internal energy, we can calculate the changed volume by the following formula:
[tex]Energy\ Supplied = Work = W = P \Delta V \\W = P(V_{f}-V_{i})\\[/tex]
where,
W = Work = Energy Supplied = 1000 J
P = Pressure = 101325 Pa
Vf = Final Volume = ?
Vi = Initial Volume = 0.001 m³
Therefore,
[tex]1000\ J = (101325\ Pa)(V_{f}-0.001\ m^3)\\\\V_{f}-0.001\ m^3 = \frac{1000\ J}{101325\ Pa}\\\\V_{f} = 0.001\ m^3 + 0.0099\ m^3[/tex]
Vf = 0.0108 m³
The force of friction depends upon
Answer:
ask internet?
Explanation:
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].
Currently, accepted theories DO NOT predict single existing particles having electrical charge equal *
a.half elementary charge
b.one elementary charge
c.two elementary charges
d.zero elementary charge
Answer
A. Half elementary charge
Explanation:
what is the magnitude of electric field at apoint 2 meter from a point charge q= 4nc
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
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
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
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:
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.
what the importance of paying orbit to the earth.
Answer:
The what?
Explanation:
When will heat STOP flowing
between two objects?
A. When one object is significantly
warmer than the other
B. When one object is significantly cooler
than the other
C. Never
D. When both objects reach the same
temperature
Answer: D. When both objects reach the same temperature.
Explanation: When will heat flow between the objects stop? Heat will always flow from the warmer object to the colder object. The heat transfer will stop when the two objects are at the same temperature and reach thermal equilibrium.
Answer: Heat will always flow from the warmer object to the colder object. The heat transfer will stop when the two objects are at the same temperature and reach thermal equilibrium.
so the answer to your question is c
Explanation:
Which is a primary energy source used by power plants to generate electricity?
coal
O wood
o gasoline
batteries
Answer:
Your answer is wood that is what they used
The primary energy source used by power plants to generate electricity is coal.
What are power plants?Power plants are plants which use fuels to generate electricity for use by homes and industries.
Power plants have different energy sources for their fuel.
The primary energy source used by power plants to generate electricity is coal.
Therefore, coal is a primary source of fuel for power plants.
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3. A backpack weighs 8.2 N and has a mass of 5 kg on the moon. What is t
strength of gravity on the moon?
8.2 N
A child swings with a small amplitude on a playground with a 2.5m long chain.
a) what is the period of the child’s motion
b) what is the frequency of vibration
a. The period of the child's motion is 3.171 seconds.
b. The frequency of the vibration is 0.315 Hz.
Calculation of the period & frequency:Since A child swings with a small amplitude on a playground with a 2.5m long chain.
a. So here the time period should be
[tex]2\times \ pi(L/g)^{1/2}\\\\2\times 3.14(2.5/9.81)^{1/2}[/tex]
T=3.171s
b) Now the frequency is
[tex]T=1\div f\\\\f=1\div 3.171[/tex]
f=0.315Hz
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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