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VARVARA [1.3K]
3 years ago
7

A major artery with a 1.7 cm2 cross-sectional area branches into 18 smaller arteries, each with an average cross-sectional area

of 0.4 cm2. By what factor is the average velocity of the blood reduced when it passes into these branches
Physics
1 answer:
Alex3 years ago
8 0

Answer:

0.14

Explanation:

Flow rate is the volume flowing through a point at a particular time, in calcuing flow rate we have

Q= v*t

it in terms of Area, we have Q= A*v

Where A= area

v= velocity.

Solving the question , flow rate is constant then

A*v= constant

A(i) v(i)= A(f) v(f)

Where A(i)= initial area= 1.00cm^2

A(f)= final area= 0.400cm^2

V(i) and V(f) are the initial and final velocity respectively and the ratio of the two will gives us the factor

Substitute the values into the equation we have

1 V(i)= 4 V(f)

But we were told that the cross sectional area of 1.00cm^2 branches into 18 smaller arteries.

Then

1 V(i)=0.4 V(f)*(18)

1 V(i)=7.2V(f)

Then if we find the ratio of the velocity, we will get the factor.

V(f)/V(i)= 1/7.2

V(f)/V(i)=0.14

Hence, the factor of the average velocity of the blood reduced when it passes into these branches is 0.14

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alexandr402 [8]

Answer:

6.5646\times 10^{16}~ \text{Hz}

Explanation:

\text{Frequency,}~ f= \dfrac{c}{\lambda} = \dfrac{3\times 10^8 }{4.57 \times 10^{-9}} = 6.5646\times 10^{16}~ \text{Hz}

5 0
3 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

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Answer:

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Over the time interval after a difference in potential is applied between the ends of a wire, what would happen to the drift vel
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The the drift velocity of the electrons is determined by atom vibrations in the crystal lattice.

<h3>How to explain the information?</h3>

Assume we could increase the average time between collisions in a typical metal to get to a limit of zero resistance. The free electrons would therefore be continuously accelerated by a constant applied voltage, according to the classical paradigm of conduction. Both the current and the drift speed would gradually pick up over time.

Although it is not the scenario implied by the question, it is possible to switch to zero resistance by using a superconducting wire instead of the usual metal. In this scenario, the maximum current is constrained, the drift velocity of the electrons is determined by atom vibrations in the crystal lattice, and it is difficult to produce a potential difference across the superconductor.

Learn more about electrons in:

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