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aalyn [17]
3 years ago
13

An intravenous (IV) system is supplying saline solution to a patient at the rate of 0.06 cm3/s through a needle of radius 0.2 mm

and length 6.32 cm.
What gauge pressure (in Pa) is needed at the entrance of the needle to cause this flow?
Assume that the viscosity of the saline solution to be the same as that of water, η = 1.0*10-3 Pa-s, and that the gauge pressure of the blood in the vein is 1500 Pa.

Enter an integer
Physics
1 answer:
horsena [70]3 years ago
8 0

Answer:

Pressure applied to the needle is 7528 Pa

Explanation:

As we know by poiseuille's law of flow of liquid through a cylindrical pipe

the rate of flow through the pipe is given as

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

now we know that

Q = 0.06 \times 10^{-6} m^3/s

radius = 0.2 mm

Length = 6.32 cm

\eta = 1\times 10^{-3} Pa s

now we have

6 \times 10^{-8} = \frac{\Delta P \pi (0.2 \times 10^{-3})^4}{8(1 \times 10^{-3})6.32 \times 10^{-2}}

3.03 \times 10^{-11} = \Delta P 5.02 \times 10^{-15}

\Delta P = 6028 Pa

now we have

P - 1500 = 6028 Pa

P = 7528 Pa

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What is the voltage across each resistor?
kogti [31]

Answer:

see solution below

Explanation:

The given resistors are connected in series.

Equivalent resistance in series = 30 + 55 + 15

Equivalent resistance in series Rt = 100 ohms

Since the potential difference in the circuit = 36V

Get the current in the circuit first

I = V/Rt

I = 36/100

I = 0.36A

Get the voltage across 30ohms resistor;

V30 = 0.36 * 30

V30  = 10.8volts

Hence the voltage across the 30ohms resistor is 10.8volts

Get the voltage across 55ohms resistor;

V55 = 0.36 * 55

V55  = 19.8volts

Hence the voltage across the 55ohms resistor is 19.8volts

Get the voltage across 15ohms resistor;

V15 = 0.36 * 15

V15  = 5.4volts

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When a magnetic field is first turned on, the flux through a 20-turn loop varies with time according to Φm=5.0t2−2.0t, where Φm
goldfiish [28.3K]

Answer:

A) ( - 200t + 40 ) volts

B)  b) anticlockwise ,  c) anticlockwise , d) clockwise ,  e) clockwise

Explanation:

Given data:

magnetic flux (Φm) = 5.0t^2 − 2.0t

number of turns = 20

<u>a) determine induced emf </u>

E = - N \frac{d\beta }{dt}

  =  - N ( 10t - 2 ) = - 20 ( 10t - 2 )

  =  - 200t + 40  volts

<u>b) Determine direction of induced current </u>

i) at t = 0

 E = - 0 + 40  ( anticlockwise direction )

ii) at t = 0.10

E = -20 + 40 =  20 ( anticlockwise direction )

iii) at t = 1

E = - 200 + 40 = - 160 ( clockwise direction)

iv) at t = 2

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8 0
3 years ago
What would be the escape speed for a craft launched from a space elevator at a height of 54,000 km?
Natasha_Volkova [10]

Answer: 3.63 km/s

Explanation:

The escape velocity equation for a craft launched from the Earth surface is:

V_{e}=\sqrt{\frac{2GM}{R}}

Where:

V_{e} is the escape velocity

G=6.67(10)^{-11} Nm^{2}/kg^{2} is the Universal Gravitational constant

M=5.976(10)^{24}kg is the mass of the Earth

R=6371 km=6371000 m is the Earth's radius

However, in this situation the craft would be launched at a height h=54000 km=54000000 m over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

V_{e}=\sqrt{\frac{2GM}{R+h}}

V_{e}=\sqrt{\frac{2(6.67(10)^{-11} Nm^{2}/kg^{2})(5.976(10)^{24}kg)}{6371000 m+54000000 m}}

Finally:

V_{e}=3633.86 m/s \approx 3.63 km/s

7 0
3 years ago
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