Answer:
it would be 83% in a fatal crash.
Your physics teachers want to race each other using wagon rockets.
The fastest wagon accelerates at 4 ms-2.
Calculate the distance traveled in 10 seconds, if it starts from rest.
Answer:
s=1/2at^2
s=1/2 x 4 x(10)^2=200m
As the length of a wire increases, the resistance
a. increases.
b. decreases.
c. remains the same.
d. disappears.
Answer:
a. increases
Explanation:
the law says
R=ρ[tex]\frac{L}{A}[/tex]
R= resistance
ρ= permiability of wire
L= length of the wire
A= area of the wire
What year did Badminton become a full-medal Olympic sport?
Answer:
1992
Explanation:
Badminton made its debut as a demonstration sport at the 1972 Olympic Games in Munich. It was not until the 1992 Games in Barcelona that it was officially included on the Olympic programme, with men's and women's singles and doubles events.
Please help
A 90 kg football player is running at 0.9 m/s. Assuming a head on collision, at what speed would a 110 kg football player have to run into the 90 kg
player in order to stop him
Answer:
The two moments must be the same:
p1=p2
m1v1=m2v2
v2=(m1v1)/m2
v2=(90 kg x 0.9 m/s)/110kg=0.7 m/s
A capacitor with air between its plates ischarged to 60 V and then disconnected fromthe battery. When a piece of glass is placedbetween the plates, the voltage across thecapacitor drops to 46 V.What is the dielectric constant of this glass?Assume the glass completely fills the spacebetween the plates.
Answer:
k = 1.30
Explanation:
For this exercise let's write the capacitance in air and with dielectric
air C₀ = Q / DV
dielectric C = k Q / DV
They tell us that the capacitor is charged and then the battery is disconnected, therefore the charge stored on the plate remains constant.
therefore the capacitance a changes to the value
C = k C₀
The voltage in the presence of dielectric must meet the relationship
ΔV = ΔV₀ / k
k = ΔV₀ /ΔV
let's calculate
k = 60/46
k = 1.30
Changes of
Which now represents the change of state described
above?
M
ham shows changes of state between solid.
se atoms of a substance lase energy
Bacharest Betore the change, the atoms
ether but are able to slide past one another
Answer:b yuh
Explanation:
Yuh
A baseball is thrown a distance of 20 m what is its speed if it takes 0.5 seconds to cover the distance
Answer:
40m/s
Explanation:
A 1.6 kg ball is attached to the end of a 0.40 m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.80 kg block that is at rest on a horizontal frictionless surface. The speed of the block just after the collision is 3 m/s. What is the speed of the ball just after the collision
Answer:
the speed of the ball just after the collision is 1.5 m/s.
Explanation:
Given;
mass of the ball, m₁ = 1.6 kg
initial velocity of the ball, u₁ = 0
mass of the block, m₂ = 0.8 kg
initial velocity of the block, u₂ = 0
final velocity of the block, v₂ = 3 m/s
let the final velocity of the ball after collision = v₁
Apply the principle of conservation of linear momentum for elastic collision;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
1.6 x 0 + 0.8 x 0 = 1.6 x v₁ + 0.8 x 3
0 = 1.6v₁ + 2.4
-1.6v₁ = 2.4
v₁ = -2.4 / 1.6
v₁ = - 1.5 m/s
v₁ = 1.5 m/s (in opposite direction of the block)
Therefore, the speed of the ball just after the collision is 1.5 m/s.
Young's double slit experiment is one of the quintessential experiments in physics. The availability of low cost lasers in recent years allows us to perform the double slit experiment rather easily in class. Your professor shines a green laser (560 nm) on a double slit with a separation of 0.108 mm. The diffraction pattern shines on the classroom wall 3.0 m away. Calculate the fringe separation between the third order and central fringe.
Answer:
y = 4.666 10⁻² m
Explanation:
The constructive interference experiment for the double slit
d sin sin θ = m λ
Let's use trigonometry to find a sine relationship.
Tan θ = y / L
tan θ = sin θ/ cos θ
in these experiments the angles are very small
tan θ = sin θ
sin θ = y / L
[tex]d \frac{y}{L}[/tex] = m λ
y = [tex]\frac{ m \lambda \ L}{d}[/tex]
we replace the values
y = 3 560 10⁻⁹ 3.0 / 0.108 10⁻³
y = 4.666 10⁻² m
What happens when a tennis racket hits
a ball?
A. The ball pushes back on the racket in the opposite
direction.
B. The ball pushes back on the racket in the same
direction.
C. The ball does not push back on the racket.
D. The ball pushes back on the racket perpendicularly.
Newton's third law of motion states that for every action, there is an equal and opposite reaction. When a tennis racket hits a ball A. The ball pushes back on the racket in the opposite direction.
When a tennis racket hits a ball, the ball exerts a force on the racket, and according to Newton's third law of motion, the racket exerts an equal and opposite force on the ball. This means that the ball pushes back on the racket in the opposite direction to which the racket struck the ball.
This principle is often referred to as "action-reaction" or "equal and opposite forces." When the racket collides with the ball, the force applied by the racket causes the ball to accelerate in the opposite direction, leading to its movement away from the racket.
Therefore, when a tennis racket hits a ball A. The ball pushes back on the racket in the opposite direction.
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Which types of reactions are also redox reactions?
Answer:
Your ans is B)
Explanation:
Hope its right
if wrong firgive me
Coulomb is the unit of which quantity?
Answer:
electric charge
Answer:
Electric charge.
Explanation:
Coulomb, unit of electric charge in the metre - kilogram - second - ampere system, the basis of the SI system of physical units. It is abbreviated as C. The coulomb is defined as the quantity of electricity transported in one second by a current of one ampere.
What is the number of waves that pass a fixed ping in a given amount of time?
A frictionless piston-cylinder contains carbon dioxide gas (CO2) initially at 500oC and 2 MPa. The system is o3 cooled in an isobaric process until the final temperature becomes 350 C and the final volume is 1 m . For this process, determine: (a) The reduced pressure and the reduced temperature of the initial state. (b) The initial volume of the piston-cylinder (in m3). (c) The mass of CO2 in the piston-cylinder (in kg). (d) The total boundary work for the process (in kJ). (e) The amount of heat transfer during the cooling process (in kJ).
Answer:
(a) The reduced pressure is 0.2711 MPa
The reduced temperature is 2.54 K
(b) The initial volume of the piston is approximately 0.806 m³
(c) The mass of CO₂ is approximately 16.9884 kg
(d) The work done, W is approximately 388.023 kJ
(e) The heat transfer is approximately -2,650.1904 kJ
Explanation:
The initial temperature of the piston-cylinder, T₁ = 500°C = 773.15 K
The initial pressure of the gas, P₁ = 2 MPa
The final temperature of the gas, T₂ = 350°C
The final volume of the gas = 1 m³
(a) For an isobaric process, we have;
The reduced pressure,
[tex]P_r = \dfrac{P}{P_c}[/tex]
The critical pressure of carbon dioxide, [tex]P_c[/tex] = 7.3773 MPa
[tex]P_r = \dfrac{2 \, MPa}{7.3773 \, MPa} \approx 0.2711 \, MPa[/tex]
The reduced pressure, [tex]P_r[/tex] = 0.2711 MPa
The critical temperature, [tex]T_c[/tex] = 304.13 K
The reduced temperature, [tex]T_r[/tex], is given by the following formula;
[tex]T_r = \dfrac{T}{T_c}[/tex]
Therefore, [tex]T_r[/tex] = (773.15 K)/(304.13 K) = 2.54216947 K
The reduced temperature, [tex]T_r[/tex] ≈ 2.54 K
(b) The initial volume of the piston, V₁ = (V₂/T₂) × T₁
∴ V₁ = (1 m³/773.15) × 623.15 = 0.80598848865 m³ ≈ 0.806 m³
The initial volume of the piston, V₁ ≈ 0.806 m³
(c) The number of moles of CO₂ in the cylinder, 'n', is given according to the following formula;
n = P·V/(T·R)
The universal gas constant, n = (2 × 10⁶Pa × 1 m³)/(623.15 K × 8.3145 J/(mol·K)) ≈ 386.0124 moles
The mass of CO₂ ≈ 386.0124 moles × 44.01 g/mol = 16.9884 kg
(d) The work done, W = P·([tex]V_f - V_i[/tex])
W = 2 × 10⁶ × (1 - 0.80598848865) = 388023.0227
The work done, W ≈ 388.023 kJ
(e) The heat transfer dQ = m·[tex]c_p[/tex] ×(T₂ - T₁)
[tex]c_p[/tex] for CO₂ ≈ 1.04 kJ/(kg·K)
∴ dQ = 16.9884 × 1.04 × (350 - 500) = -2,650.1904 kJ
Therefore, the heat transfer = dQ = -2,650.1904 kJ
Explain why the sound waves always reach the observer after the light waves
Answer:
I think it's because the light waves travel faster than the sound waves.
The speed of light is far greater than the speed of sound hence, sound waves always reach the observer after the light waves.
What is the speed of light?Light occurs in the electromagnetic spectrum. Recall that light can be transmitted through vaccuum unlike sound.
The speed of light is far greater than the speed of sound hence, sound waves always reach the observer after the light waves.
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Given that the collision is elastic and glider 2 is initially at rest (v2,i =0), please use below Eqs. to explain why
1. Glider 2 will be always kicked toward the same direction as glider 1 comes in (v2,f and v1,i have the same sign)
2. Glider 1 will bounce back (v1f and v1,i have opposite sign) if it is lighter than glider 2 (m1= m2).
3. Glider 1 will keep moving forward (v1,f and v1 i have the same sign) if it is heavier than glider 2 (m1 m2)
4. Glider 1 will stop (v1f =0) if it weighs the same as glider 2 (m1= m2)
v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2
v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2
Answer:
Explanation:
1 )
Put v2,i =0, in second equation
v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2
v2,f = 0 + 2m1v1,i/m1+m2
v2,f = 2m1v1,i/m1+m2
In this equation coefficient of v1,i is positive so v2,f and v1,i have the same sign.
2 )
Put m1 < m2 and v2,i =0 in first equation
v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2
v1,f= (m1-m2)v1,i
As m1-m2 is negative , v1f and v1i will have opposite sign.
3 )
Put m1 > m2 and v2,i =0 in first equation
v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2
v1,f= (m1-m2)v1,i
m1 - m2 is positive so v1f and v1i will have same sign.
4 )
Put m1 = m2 and v2,i =0 in first equation
v1,f= (m1-m2)v1,i
= 0 because m1 = m2
So glider 1 will stop because v1,f = 0 .
When the glider 2 is initially at rest (v2,i =0) and collision is elastic, the different equation are explained for different conditions of glider 1.
What is conservation of momentum?Momentum of a object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity.
When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.
The given equations are,
[tex]v_{1f}= \dfrac{(m_1-m_2)v_{1i} + 2m_2v_{2i}}{m1+m2}[/tex]
[tex]v_{2f}= \dfrac{(m_2-m_1)v_{2i} + 2m_1v_{1i}}{m1+m2}[/tex]
Given that the collision is elastic and glider 2 is initially at rest (v2,i =0),
1. Glider 2 will always be kicked toward the same direction as glider 1 comes in (v2,f and v1,i have the same sign)Put [tex]v_{2i}[/tex] equal to zero, in the given equation as,
[tex]v_{2f}= \dfrac{(m_2-m_1)(0) + 2m_1v_{1i}}{m1+m2}\\v_{2f}= \dfrac{ 2m_1v_{1i}}{m1+m2}[/tex]
v2,f and v1,i have the same sign in the above equation.
2. Glider 1 will bounce back (v1f and v1,i have opposite sign) if it is lighter than glider 2 (m1<m2).Put [tex]v_{1i}[/tex] equal to zero, in the given equation as,
[tex]v_{1f}= \dfrac{(m_2-m_1)(0) + 2m_2v_{2i}}{m1+m2}\\v_{1f}= \dfrac{ 2m_2v_{2i}}{m1+m2}[/tex]
Now, if we consider (m1<m2), for the above equation, the result will be negative. Thus, v1f and v1,i have opposite sign.
3. Glider 1 will keep moving forward (v1,f and v1 i have the same sign) if it is heavier than glider 2 (m1 >m2)Put [tex]v_{2i}[/tex] equal to zero, in the first equation as,
[tex]v_{1f}= \dfrac{(m_2-m_1)(v_{1i}) + 2m_2(0)}{m1+m2}\\v_{1f}= \dfrac{(m_2-m_1)(v_{1i})}{m1+m2}[/tex]
Now, if we consider (m1>m2), for the above equation, the result will be positive. v1,f and v1 i have the same sign.
4. Glider 1 will stop (v1f =0) if it weighs the same as glider 2 (m1= m2)Put [tex]v_{2i}[/tex] equal to zero, in the first equation as,
[tex]v_{1f}= \dfrac{(m_2-m_1)(v_{1i}) + 2m_2(0)}{m1+m2}\\v_{1f}= \dfrac{(m_2-m_1)(v_{1i})}{m1+m2}[/tex]
Now, if we consider (m1=m2), for the above equation, the result will be zero. Thus the glider 1 will stop (v1f =0).
Hence, when the glider 2 is initially at rest (v2,i =0) and collision is elastic, the different equation are explained for different conditions of glider 1.
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Three cylindrical wires, 1, 2, and 3 are made of the same materialand have resistances R1, R2, and R3, respectively. Wires 1 and 2 have the same lengthbut diameter of wire 2 is twice that of wire 1. Wires 2 and 3 have the same diameterbut length of wire 3is twice that of wire 2.
1. Rank the wires according to their resistances, greatest first.
A. R1> R2> R3R1
B. R1> R3> R2R2
C. R2> R1> R3R3
D. R2> R3> R1
E. R3> R1> R2
F. R3> R2> R1Q2.
2. If same voltage is applied across each of the wires, which one will dissipate heat at the highest rate
Answer:
1) R₁ > R₃ > R₂ correct B , 2) the wire that dissipates the most is wire 2
Explanation:
1) The resistance of a wire is given by the expression
R = [tex]\rho \ \frac{l}{A}[/tex]
where ρ is the resistivity of the material, l the length of the wire and A the area of the wire
The area is given by
A = π r² = π d² / 4
we substitute
R = (ρ 4 /π) [tex]\frac{l}{d^2}[/tex]
the amount in parentheses is constant for this case
let's analyze the situation presented, to find the resistance of each wire
* indicate l₁ = l₂ and d₂ = 2 d₁
the resistance of wire 1 is
R₁ = (ρ 4 /π) [tex]\frac{l_1}{d_1^2}[/tex]
the resistance of wire 2 is
R₂ = (ρ 4 /π) \frac{l_2}{d_2^2}
R₂ = (ρ 4 /π) [tex]\frac{l_1}{ (2 d_1)^2}[/tex]
R₂ = (ρ 4 /π ) [tex]\frac{l_1}{d_1^2}[/tex] ¼
R₂ = ¼ R₁
* indicate that d₂ = d₃ and l₃ = 2 l₂
R2 = (ρ 4 /π) [tex]\frac{l_2}{d_2^2}[/tex]
the resistance of wire 3 is substituting the indicated condition
R3 = (ρ 4 /π 2) \frac{l_3}{d_3^2}
R3 = (ρ 4 /π) [tex]\frac{2 \ l_2}{d_2}[/tex]
R3 = 2 R₂
let's write the relations obtained
R₁ = (ρ 4 /π) [tex]\frac{I_1}{ d_1^2}[/tex]
R₂ = ¼ R₁
R₃ = 2 R₂
let's write everything as a function of R1
R₁ =(ρ 4 /π) [tex]\frac{l_1}{d_1^2}[/tex]
R₂ = ¼ R₁
R₃ = ½ R₁
the resistance of the wire in decreasing order is
R₁ > R₃ > R₂
2) The power dissipated by a wire is
P = V I
the voltage is
V = I R
I = V / R
substituting
P = V² / R
therefore the power dissipated by each wire is
wire 1
P₁ = V² / R₁
wire 2
P₂ = V² / R₂
P₂ = [tex]\frac{V^2}{ \frac{1}{4} R_1}[/tex]
P₂ = 4 P₁
wire 3
P₃ = V² / R₃
P₃ = [tex]\frac{V^2}{ \frac{1}{2} R_1}[/tex]
P₃ = 2 P₁
Therefore, the wire that dissipates the most is wire 2
A 65 kg cart travels at a constant speed of 4.6 m/s. What is its kinetic energy?
Answer:
Explanation:
mass (m) = 65 kg
velocity (v) = 4.6 m/s
Kinetic energy (KE)
= 1/2 * m * v²
= 1/2 * 65 * 4.6²
= 687.7 J
hope it helps :)
How much is Coulomb's constant ?
Answer:
8.9875517923(14)×109 kg⋅m3⋅s−2⋅C−2
Explanation:
Exact number is 8.9875517923(14)×109 kg⋅m3⋅s−2⋅C−2
Which bond(s) shown are double bonds?
Answer:
in the diagram shown, Oxygen is the only one that has a double bond.
Explanation:
Chemical bonds are the way that electors are shared between atoms to form molecules whose energy is less than the energy of individual atoms.
A graphical way to show these electrons is to represent them by point in the atoms. Every other point represents a shared bond or pair of electrons.
Therefore for simple bonds two electrons are drawn. For double bonds four electrons and for triple bonds six electrons.
Therefore, in the diagram shown, Oxygen is the only one that has a double bond.
how could you increase the gravitational potential energy of an object without changing its mass and gravity
Create a concept map that shows the relationships among the following: crest, trough, compression, rarefaction, wave length, wave frequency, amplitude, and wave
speed.
Uhm.. based on what? Can you please specify/add an image or some more text and elaborate?
A 50kg bicyclist on a 10kg bicycle speeds up from 5.0m/s to 10m/s. What was the total kinetic energy after accelerating?
Answer:
Explanation:
Ek=1/2mv^2
m=50+10=60,v=10
Ek=1/2*60*100=3000J
After the first trial your sibling spills juice on the track when you run the car down the track again you measure the speed at the bottom of the track to be 2.0 MS how much of the total mechanical energy was converted to thermal energy as a car moved through the sticky juice?
Plz help asap
What is the first velocity of the car with three washers
at the 0.25 meter mark?
m/s
Calculate the first and second velocities of the car with
three washers attached to the pulley, using the formulas
V1 = 0.25 m/ty, and
v2 = 0.25 m/(t2- ty)
where t1 and t2 are the average times the car took to
reach the 0.25 and the 0.50 meter marks. Record these
velocities, to two decimal places, in Table E.
What is the second velocity of the car with three
washers at the 0.50 meter mark?
m/s
Answer:
What is the first velocity of the car with three washers
at the 0.25 meter mark? 0.19
What is the second velocity of the car with three
washers at the 0.50 meter mark? 0.45
Explanation:
Look at the question carefully.
The person above me got the answer wrong.
That answer is for a very similar question, but not this one.
(1) The first velocity of the car with two washers at the 0.25 meter mark is 0.125 m/s.
(2) The second velocity of the car with two washers at the 0.5 meter mark is 0.25 m/s.
What is velocity?Velocity is the rate of change of displacement with time. The velocity of the cars depends on displacement and time of motion.
First velocity of the car at the 0.25 meter mark
The first velocity is calculated as follows;
v1 = 0.25 m / t1
let t₁ = 2 s
v₁ = 0.25/2
v₁ = 0.125 m/s
Second velocity of the car at the 0.5 meter mark
v2 = 0.25 m / (t2 – t1)
let t₂ = 3 s
v₂ = 0.25(3 - 2)
v₂ = 0.25 m/s
The ratio of distance covered by an object in a specific direction and the time taken to cover the distance is known as the velocity of the object. Mathematically, the expression for the velocity is,
v = d/t
Here, d is the distance covered.
And t is the average time taken to cover the distance.
Then the first velocity of the car at 0.25 m is,
v1 = d1/t1
v1 = 0.25 / t1
Here, t1 is the average time for the first distance.
And the second velocity of the car with four washers at the 0. 50 m mark is,
v2 = d2/t2
v2 = 0.50 /t2
Therefore, here, t2 is the average time for the second distance.
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strong bases are good conductors of
Answer:
Electricity ⚡️
Explanation:
Two masses, each having a value of M, are vibrating vertically on a spring with a Hooke's law constant, k. At the lowest point of the vibration, one of the masses falls off, so that now the total mass is M instead of 2M. Comparing the new vibrational motion to the original vibrational motion: 1) How is the period of vibration different, if at all
An automobile tire with a radius of 0.30 m starts at rest and accelerates at a constant angular acceleration of 2.0 rad/s^2 for 7.0 s. What is the angular displacement of the tire?
A. 32 rad
B. 25 rad
C. 28 rad
D. 49 rad
The tire undergoes an angular displacement θ in time t according to
θ = θ₀ + ω₀ t + 1/2 α t ²
where
θ₀ = initial angular displacement
ω₀ = init. ang. velocity
α = ang. acceleration
Here we have θ₀ = 0 and ω₀, so after t = 7.0 s with acceleration α = 2.0 rad/s², the wheel will have been displaced
θ = 1/2 (2.0 rad/s²) (7.0 s)² = 49 rad
making the answer D.
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
While eating a peanut butter and jelly sandwich, peanut butter drips on Diana’s shirt. She tries to remove the peanut butter from her shirt by rubbing the spot with water, but it doesn’t work. What is the best explanation for why the water does not remove the peanut butter?
The water and peanut butter have different chemical properties.
The water and peanut butter have different temperatures.
The water and peanut butter have different pressures.
Answer:
The water and peanut butter have different chemical properties.
(Lipids are not soluble in water)
Answer:
a. The water and peanut butter have different chemical properties.
Explanation: