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Gnoma [55]
3 years ago
6

An artery with cross sectional area of 1 cm2 branches into 20 smaller arteries each with 0.5 cm2 cross sectional area. If the ve

locity of blood in the thicker artery is v, what is the velocity of the blood in the thinner arteries
Physics
1 answer:
kow [346]3 years ago
4 0

Answer:

The velocity of the blood in the thinner arteries is 0.1 times that of the thicker artery.

Explanation:

To find the velocity of the blood we need to use the continuity equation:

n_{1}A_{1}v_{1} = n_{2}A_{2}v_{2}   (1)  

<u>Where</u>:

n: is the number of branches

A: is the cross-sectional area

v: is the velocity

For artery 1, we have:

n₁ = 1, A₁ = 1 cm², v₁ = v

For the 20 arteries (2), we have:

n₂ = 20, A₂ = 0.5 cm², v₂ =?

By using equation (1):

n_{1}A_{1}v_{1} = n_{2}A_{2}v_{2}

1 cm^{2}*v = 20*0.5 cm^{2}*v_{2}

v_{2} = \frac{1 cm^{2}*v}{20*0.5 cm^{2}} = \frac{v}{10} = 0.1v

Therefore, the velocity of the blood in the thinner arteries is 0.1 times that of the thicker artery.

I hope it helps you!      

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A psychopathologist records the number of criminal offenses among teenage drug users in a nationwide sample of 1,201 participant
lara [203]

Answer:

Standard deviation = 3

Explanation:

Given

N = 1201

SS = 10800

Required

Determine the standard deviation

First, we need to determine the variance;

Variance = \frac{SS}{N - 1}

This gives:

Variance = \frac{10800}{1201 - 1}

Variance = \frac{10800}{1200}

Variance = 9

Know that:

Variance = SD^2

Where SD represents standard deviation

This gives

9 = SD^2

Take square root

SD = \sqrt 9

SD = 3

7 0
3 years ago
Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t
n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

y=\frac{\lambda (m+\frac{1}{2})D}{d}

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

\lambda is the wavelength of the light used

In this problem we have:

\lambda=683 nm = 683\cdot 10^{-9} m is the wavelength of the light

d=1.1 mm = 0.0011 m is the width of the slit

m = 13 is the order of the minimum

y=8.57 cm = 0.0857 m is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m

6 0
3 years ago
What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t
Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

\Delta P =  \frac{8 \mu L Q}{\pi r^4}

where:

\Delta P is the pressure difference between the two ends

\mu is viscosity of the fluid

L is the length of the pipe

Q=Av is the volumetric flow rate, with A=\pi r^2 being the section of the tube and v the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

\mu=0.001 Pa/s is the dynamic water viscosity at 20^{\circ}

L=4.0 cm=0.04 m

Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s

and r=1 mm=0.001 m

Using these data in the formula, we get:

\Delta P = 3200 Pa

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0
3 years ago
The surface area of a postage
ANTONII [103]

Answer:

608

Explanation:

Trust me

4 0
2 years ago
Read 2 more answers
Air flows through an adiabatic turbine that is in steady operation. The air enters at 150 psia, 900oF, and 350 ft/s and leaves a
Nonamiya [84]

Answer:

1486.5\frac{Btu}{s}

Explanation:

The inlet specific volume of air is given by:

v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{(0.3704\frac{psia.ft^3}{lbm.R})(1360R)}{150psia}\\\\v_1=3.358\frac{ft^3}{lbm} \ \ \ \  \ \  \ \ \...i

The mass flow rates is expressed as:

\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}

The energy balance for the system can the be expresses in the rate form as:

E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}

Hence, the mass flow rate of the air is 1486.5Btu/s

5 0
3 years ago
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