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Crazy boy [7]
4 years ago
15

A major artery with a 1.6 cm2 cross-sectional area branches into 18 smaller arteries, each with an average cross-sectional area

of 0.7 cm2. By what factor is the average velocity of the blood reduced when it passes into these branches?
Physics
1 answer:
Brums [2.3K]4 years ago
5 0

To solve this problem it is necessary to apply the continuity equations for which it is defined that the proportion of Area in the initial section is equal to the final section. In other words,

A_1 v_1 = A_2 v_2

Where,

A_i = Cross sectional area at each section

v_i =Velocities of fluid at each section

The total area of the branch is eighteen times of area of smaller artery. The average cross-sectional area of each artery is 0.7cm^2.

Therefore the Cross-sectional area at the end is

A_2= 18*0.7cm^2

A_2 = 12.6cm^2

Applying the previous equation we have then

A_1 v_1 = A_2 v_2

(1.7cm^2) v_1 = (12.6cm^2)v_2

The ratio of the velocities then is

\frac{v_1}{v_2} = \frac{1.7}{12.6}

\frac{v_1}{v_2} = 0.135

Therefore the factor by which the velocity of blood will reduce when it enters the smaller arteries is 0.1349

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Find the volume v of the solid obtained by rotating the region bounded by the given curves about the specified line. y = x2, y =
Simora [160]

The correct answer for  the volume v of the solid obtained by rotating the region bounded by the given curves about the specified line. y = x2, y = 4x is 128/3 * π.

Volume of Solid of Revolution by Shell method is given by

V = 2π * integrate x(height) dx Here, height = 4x-x2

(1)& x-varies from x = 0 to x = 4 then from eqn(1)  V = 2π * integrate x(4x - x ^ 2) dx from x = 0 to 4 = 2π * integrate (4x ^ 2 - x ^ 3) dx from x = 0 to 4

Basic Rule(1) ∫ x^n dx =x^ n+1/ n+1

V=2 π [4((x ^ 3)/3) - (x ^ 4)/4] 0 ^ 4 =2 π[ 4/3 x^ 3 - x^ 4/4 ] 0 ^ 4

V = 2π [4/3 * 4 ^ 3 - (4 ^ 4)/4} - 0]

V = 128/3 * π.

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4 0
2 years ago
A small metal sphere has a mass of 0.14 g and a charge of -22.0 nc . it is 10 cm directly above an identical sphere with the sam
Allushta [10]
For this problem, we use the Coulomb's law written in equation as:

F = kQ₁Q₂/d²
where
F is the electrical force
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Substituting the values:

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F = 0.0004356 N
4 0
3 years ago
Which stretch focuses on stretching the glutes and hamstrings while laying on your back?
WINSTONCH [101]

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Explanation:

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3 years ago
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A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif
Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c)  w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)  

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

                    μ = N*I*A

Where,

           N: The number of turns

           I : Current in coil

           A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

                    μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

                    μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

             vec ( τi ) = vec ( μi ) x vec ( B )

              = 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

              = ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

          \left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right]  = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\

- The initial torque ( τi ) is written as follows:

           vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

           

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

          Ui = - vec ( μi ) . vec ( B )

          Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

          Ui = - [(-0.000605556 + 0.00011)]

          Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

          Uf = - vec ( μf ) . vec ( B )

          Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

          Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

          ΔU = Uf - Ui

          ΔU = -0.000252315 - 0.000495556  

          ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

                Ui + Eki = Uf + Ekf

Where,

             Eki : The initial kinetic energy ( initially at rest ) = 0

             Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.    

               -ΔU = Ekf

                0.5*T*w^2 = -ΔU

                w^2 = -ΔU*2 / T

Where,

                w: The angular speed at second position

               w = √(0.000747871*2 / 6.50×10^−7)

              w = 47.97 rad / s

6 0
3 years ago
NEED HELP Two skaters stand facing each other. One skater’s mass is 60 kg, and the other’s mass is 72 kg. If the skaters push aw
stealth61 [152]

Answer:

the heavier skater has less momentum

hope it is helpful to you

3 0
3 years ago
Read 2 more answers
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