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Elis [28]
3 years ago
6

Use Poiseuille's Law to calculate the rate of flow in a small human artery where we can take η = 0.028, R = 0.008 cm, l = 2 cm,

and P = 5000 dynes/cm2. (Round your answer to three significant figures.)
Physics
1 answer:
FrozenT [24]3 years ago
8 0

Answer:

1.435 x 10^-4 cm^3/s

Explanation:

coefficient of viscosity, η = 0.028 poise

Radius, R = 0.008 cm

length, l = 2 cm

Pressure, P = 5000 dyne/cm^2

The Poiseuille's formula is given below

V=\frac{\pi PR^{4}}{8\eta l}

Where, V is the rate of flow, that means volume flowing per second

V=\frac{3.14\times5000 \times {0.008}^{4}}{8 \times 0.028 \times 2}

V = 1.435 x 10^-4 cm^3/s

Thus, the rate of flow is 1.435 x 10^-4 cm^3/s.

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A toy airplane, flying in a horizontal, circular
Goshia [24]

Answer:

8.4 m/s

Explanation:

The toy completes 10 circle in 30 seconds. So its frequency of revolution is

f=\frac{10}{30 s}=0.33 Hz

The periof of revolution is the reciprocal of the frequency, so

T=\frac{1}{f}=\frac{1}{0.33 Hz}=3 s

The radius of the circular path is

r = 4.0 m

So the total distance covered by the toy in one circle is the length of the circumference:

2\pi r

And so the average speed is

v=\frac{2\pi r}{T}=\frac{2\pi (4.0 m)}{3 s}=8.4 m/s

4 0
3 years ago
Read 2 more answers
Which statements does the Second Law of Thermodynamics support?
Mazyrski [523]
According to the second law of thermodynamics, 
the answer is 
<span>4. The entropy of the universe is increasing. </span>
4 0
3 years ago
In the standing waves experiment, the string has a mass of 31.2 g and a length of 0.7 m. The string is connected to a mechanical
mestny [16]

Answer:

linear density of the string = 4.46 × 10⁻⁴ kg/m

Explanation:

given,

mass of the string = 31.2 g

length of string = 0.7 m

linear density of the string = \dfrac{mass\ of\ string}{length}

linear density of the string = \dfrac{31.2\times 10^{-3}\ kg}{0.7\ m}

linear density of the string = 44.57 × 10⁻³ kg/m

linear density of the string = 4.46 × 10⁻⁴ kg/m

7 0
3 years ago
Fatigue strength is generally significantly improved by using high steel a. alloy b. yield c. hardened d. ultimate strength e. a
Gala2k [10]

Answer:

e. all of these

Explanation:

The fatigue strength is improved by then high alloy steels , high yield steels , high hardened steel , high ultimate steel .

Due to the formation of the improved materials in alloy steels will increase the fatigue strength . Similarly for a high yield steels and hardened steels these cycles to failure will improve .

7 0
3 years ago
A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the
Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

qvB = m\frac{v^2}{r}

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

T=\frac{2\pi r}{v}

Re-arranging for r

r=\frac{Tv}{2\pi}

And substituting into the previous equation

qvB = m \frac{Tv^3}{2\pi}

Solving for T,

T=\frac{2\pi q B}{m v^2}

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0
3 years ago
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