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polet [3.4K]
3 years ago
7

The diameter of a segment of an artery is reduced by a factor of two due to an obstruction. Assume that the flow is incompressib

le and laminar. Compared to an unobstructed segment of the artery, the velocity of blood in the obstructed segment of the artery is __________.
Physics
1 answer:
velikii [3]3 years ago
6 0

Answer:

The velocity of blood on an obstructed segment of an artery is greater than an obstructed one

Explanation:

The blood as an incompressible an laminar fluid let us to assume the flux of blood in all segments of an artery is the same, this is known as continuity equation and let us to write the change on the flux between some point 1 and some point 2 as:

\Delta Q = 0

Q_2-Q_1=0

Q_2=Q_1 (1)

So, if we take the Q1 as the flux on the unobstructed artery and Q2 the obstructed artery, we're saying the flux of blood on the obstructed artery is the same in the unobstructed one. Flux is defined as the velocity of a fluid times the cross-section area on a specific point, so (1) is:

v_2*A_2=v_1*A_1

With v the velocity and A the cross-section area, in our should be assumed as a circle of some dimeter D1 and D2 (remember the area of a circle is \frac{\pi*D^2}{4})

v_2*\frac{\pi *D_2^2}{4}=v_1*\frac{\pi *D_1^2}{4}

For an obstructed artery D2 is less than D1 and because we have an equality and \frac{\pi}{4} is constant, v2 should be greater than v1 to conserve the equality, that means the velocity of blood on an obstructed segment of an artery is greater than an obstructed one.

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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
You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp
GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d)539.55\mu

(e) 0

Explanation:

The period for 1 circle 2\pi of the merry go around is 9.5s. It means the angular speed is:

\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s

(a)The speed is

v = \omega * R = 0.661 * 17 = 11.24 m/s

(b) Centripetal acceleration:

a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2

(c) Magnitude of the force that keeps you go around at this acceleration

F = ma = 55 * 7.44 = 409 N

(d) let the coefficient of friction by \mu. The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

F_f = mg\mu = 55*9.81\mu = 539.55\mu

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3 years ago
Why is the teammate bond often so strong?
Kazeer [188]

Answer:

c

Explanation:

While people are working, they need to have a strong bond with their teammates,to face all the hardships .

thus, teammates work together, win and lose together, and often eat and live together.

3 0
3 years ago
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A cart with a mass of 120 kg and a velocity
klasskru [66]

Answer:

dumpster mass

Explanation:

you will need the mass of the dumpster to calculate using conservation of momentum

6 0
2 years ago
10 points!! If you help!!!
marta [7]

For Mass

K.E = (1/2*mv^2)

Explanation:

Kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

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