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Rudiy27
3 years ago
10

The radius of the aorta is «10 mm and the blood flowing through it has a speed of about 300 mm/s. A capillary has a radius of ab

out 4ˆ10´3 mm but there are literally billions of them. The average speed of blood through the capillaries is about 5ˆ10´4 m/s. (i) Calculate the effective cross sectional area of the capillaries and (ii) the approximate number of capillaries.
Physics
1 answer:
stealth61 [152]3 years ago
3 0

Answer:

(I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is 3.74\times10^{9}

Explanation:

Given that,

Radius of aorta = 10 mm

Speed = 300 mm/s

Radius of capillary r=4\times10^{-3}\ mm

Speed of blood v=5\times10^{-4}\ m/s

(I). We need to calculate the effective cross sectional area of the capillaries

Using continuity equation

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

Where. v₁ = speed of blood in capillarity

A₂ = area of cross section of aorta

v₂ =speed of blood in aorta

Put the value into the formula

A_{1}=A_{2}\times\dfrac{v_{2}}{v_{1}}

A_{1}=\pi\times(10\times10^{-3})^2\times\dfrac{300\times10^{-3}}{5\times10^{-4}}

A_{1}=0.188\ m^2

(II). We need to calculate the approximate number of capillaries

Using formula of area of cross section

A_{1}=N\pi r_{c}^2

N=\dfrac{A_{1}}{\pi\times r_{c}^2}

Put the value into the formula

N=\dfrac{0.188}{\pi\times(4\times10^{-6})^2}

N=3.74\times10^{9}

Hence, (I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is 3.74\times10^{9}

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Explanation: hope this helps

8 0
3 years ago
A girl coasts down a hill on a sled, reaching level ground at the bottom with a speed of 7.4 m/s. The coefficient of kinetic fri
Amanda [17]

Answer: 62.22 meters

Explanation:

Two solve this, we need to know first how the Sled interact with the snow. While the Sled runs over the snow, the snow opposes to it movement, this is call friction. Frictions creates is a force that goes against any movement and is caused by the interaction between surfaces.

Friction can be calculated using the following equation:

Friction's Force (Ff) = m*N

Where m is the Frictions coficient and N is call Normal, which is a force created by the contact between any weight and a surface. Because the sled is and the surface are totally horizontal, the Normal is equal to the Sled's Weight.

So for this case, we can calculate the Friction's Force:

Ff = 0.045 * 735 N

Ff = 33,07 N

This force will cause a negative acceleration the Sled causing it to slow down and finally stop. To know this accelaration we can use the following equation:

Force (F) = Mass (m) * Accelaration (a)

Looking for the accelaration:

Acceleration (a) = F / M

As a side note:

Mass (M) = F / a                                        (1)

Where we know that F is equal to our Friction Force (Ff) and the Mass for the girl and the Sled can be obtained using the previous equation (1) like this

Girl and sled Mass (M) = 735 N / (9.81 m/s)

M = 74,92 Kg

This is posible, because the Girl and The Sled are aplying a Force on the ground equal to 735 N. The accelartion for their mass is the gravity

Now, with the mass we can found the Accelaration caused by friction:

Acceleration (a) = Ff / M

Accelaration (a) = 33,07 N / 74,92 Kg

a = 0,44 m/s^{2}

To know how far the Sled travel we can use the following equation:

V_{f} ^{2} = V_{0} ^{2} + 2ax

Where Vf is the Final Velocity, which is when the sled finally stops (Vf = 0), Vo is the Initial Speed of the Sled equal to 7.4 m/s and a is equal to the negative accelaration that causes the friction.

(0)^{2} = (7.4 m/s)^{2} + 2(-0,44)x

Searching for x:

x = \frac{0 - (7.4 m/s)^{2} }{-0.88}

x = \frac{0 - (7.4 m/s)^{2} }{-0.88}

x = \frac{0 - (7.4 m/s)^{2} }{-0.88}

x = 62.22 meters

The girls will run over the snow 62.22 meter until finally stopping

4 0
3 years ago
1.Suppose someone pulls a cart up a ramp a distance of 85cm along the ramp with a force of 15N.
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1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

W=Fd

where

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Substituting in the formula, we get

W=(15 N)(0.85 m)=12.75 J


2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

W=Fd=12.75 J

However, in this case the displacement is

d = 120 cm = 1.20 m

Therefore, the magnitude of the force in this case is

F=\frac{W}{d}=\frac{12.75 J}{1.20 m}=10.6 N

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3 years ago
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6 0
3 years ago
The velocity of a 150 kg cart changes from 6.0 m/s to 14.0 m/s. What is the magnitude of the impulse that acted on it?
aleksandrvk [35]
M = 150 kg.

Final velocity, v = 14 m/s

Initial Velocity, u = 6 m/s

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8 0
4 years ago
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