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Ket [755]
3 years ago
10

A small artery has a length of 1.10 × 10-3 m and a radius of 2.50 × 10-5 m. If the pressure drop across the artery is 1.15 kPa,

what is the flow rate through the artery? Assume that the temperature is 37°C and the viscosity of whole blood is 2.084 × 10-3 Pa·s.
Physics
1 answer:
levacccp [35]3 years ago
8 0

Answer:

7.69533\times 10^{-11}\ m^3/s

Explanation:

P = Pressure difference = 1.15 kPa

r = Radius = 2.5\times 10^{-5}\ m

\eta = Viscosity of liquid = 2.084\times 10^{-3}\ Pas

l = Length of artery = 1.1\times 10^{-3}\ m

From Poiseuille's equation we have

Q=\frac{\pi Pr^4}{8\eta l}\\\Rightarrow Q=\frac{\pi 1.15\times 10^3\times (2.5\times 10^{-5})^4}{8\times 2.084\times 10^{-3}\times 1.1\times 10^{-3}}\\\Rightarrow Q=7.69533\times 10^{-11}\ m^3/s

The flow rate of blood is 7.69533\times 10^{-11}\ m^3/s

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Explanation:

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A diver stands on a diving platform 10.0 m above the surface of a pool and leaps upward with an initial speed of 2.5 m/s. how fa
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<span>The diver is heading downwards at 12 m/s Ignoring air resistance, the formula for the distance under constant acceleration is d = VT - 0.5AT^2 where V = initial velocity T = time A = acceleration (9.8 m/s^2 on Earth) In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m) So let's substitute the known values and solve for T d = VT - 0.5AT^2 -7 = 2.5T - 0.5*9.8T^2 -7 = 2.5T - 4.9T^2 0 = 2.5T - 4.9T^2 + 7 We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164. Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s V = 2.5 m/s - 14.47706141 m/s V = -11.97706141 m/s So the diver is going down at a velocity of 11.98 m/s Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point. V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s) V = 2.5 m/s - (-9.477061409 m/s) V = 2.5 m/s + 9.477061409 m/s V = 11.97706141 m/s And you get the exact same velocity, except it's the opposite sign. In any case, the result needs to be rounded to 2 significant figures which is -12 m/s</span>
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3 years ago
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti
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Explanation:

a=\frac{F}{m} =\frac{12N}{20kg}

a=0.6 m/s²

Vf=0=Vi-a.t

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d= Vi.t - \frac{a.t^{2}}{2}

d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m

3 0
3 years ago
Running at 1.55 m/s, Bruce, the 40.0 kg quarterback, collides with Biff, the 90.0 kg tackle, who is traveling at 7.0 m/s in the
Margarita [4]

Answer:Bruce is knocked backwards at  

14

m

s

.

Explanation:

This is a problem of momentum (

→

p

) conservation, where

→

p

=

m

→

v

and because momentum is always conserved, in a collision:

→

p

f

=

→

p

i

We are given that  

m

1

=

45

k

g

,  

v

1

=

2

m

s

,  

m

2

=

90

k

g

, and  

v

2

=

7

m

s

The momentum of Bruce (

m

1

) before the collision is given by

→

p

1

=

m

1

v

1

→

p

1

=

(

45

k

g

)

(

2

m

s

)

→

p

1

=

90

k

g

m

s

Similarly, the momentum of Biff (

m

2

) before the collision is given by

→

p

2

=

(

90

k

g

)

(

7

m

s

)

=

630

k

g

m

s

The total linear momentum before the collision is the sum of the momentums of each of the football players.

→

P

=

→

p

t

o

t

=

∑

→

p

→

P

i

=

→

p

1

+

→

p

2

→

P

i

=

90

k

g

m

s

+

630

k

g

m

s

=

720

k

g

m

s

Because momentum is conserved, we know that given a momentum of  

720

k

g

m

s

before the collision, the momentum after the collision will also be  

720

k

g

m

s

. We are given the final velocity of Biff (

v

2

=

1

m

s

) and asked to find the final velocity of Bruce.

→

P

f

=

→

p

1

f

+

→

p

2

f

→

P

f

=

m

1

v

1

f

+

m

2

v

2

f

Solve for  

v

1

:

v

1

f

=

→

P

f

−

m

2

v

2

f

m

1

Using our known values:

v

1

f

=

720

k

g

m

s

−

(

90

k

g

)

(

1

m

s

)

45

k

g

v

1

f

=

14

m

s

∴

Bruce is knocked backwards at  

14

m

s

.

Explanation:

5 0
3 years ago
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