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3 years ago

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The human circulation system has approximately 1 × 109 capillary vessels. Each vessel has a diameter of about 8 μm .Assuming the

cardiac output is 5 L/min, determine the average speed, in centimeters per second, of blood flow through each capillary vessel.

1 answer:

DerKrebs [107]3 years ago

7 0

To solve this exercise we must apply the concept of Flow as the measure given to determine the volume of a liquid flowing per unit of time, and that can be calculated through velocity and Area, mathematically this can be determined as

$\bar{v}=\frac{Q}{A}$

Q = Discharge of Flow

A = Cross sectional Area

$\bar{v} =$ Velocity

The area of the cross section of the capillary tube is

$A_c = \pi r^2$

$A_c = \pi (\frac{d}{2})^2$

$A_c = \pi (\frac{8*10^{-6}}{2})^2$

$A_c = 5.02685*10^{-11}m^2$

The total Area by this formula:

$A_1 = nA_c$

Where,

$A_c =$ Stands for area of capillary

n = Stands for number of blood vessels

$A_1 = (1*10^9)(5.0265*10^{-11})$

$A_1 = 5.0265*10^{-2}m^2$

Finally replacing at our first equation,

$\bar{v} = (\frac{5L/min}{5.0265*10^{-2}m^2})(\frac{1000cm^3}{1L})(\frac{1min}{60s})$

$\bar{v} = 1.66cm^3/s$

Therefore the average speed, in centimeters per second, of blood flow through each capillary vessel is 1.66cm^3/s

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A spherical aluminum ball of mass 1.26 kg contains an empty spherical cavity that is concentric with the ball. The ball barely f

V125BC [204]

The outer radius of the ball = 6.70cm.

<h3>What is Buoyant Force?</h3>

The upward force applied to an object that is fully or partially submerged in a fluid is known as the buoyant force.

Given: Mass of aluminum ball = $m_{b}=1.26kg$

To find: The outer radius of the ball.

Finding:

As Buoyant Force = $F_{b}=e\times(g)\times\frac{4}{3}\pi(r)^{3}$ ,

The weight of the ball should be equal to the buoyant force since it floats on the water.

That is,

=> $1.26=e_{water}(g)(\frac{4}{3}\pi(r_{outer})^{3})$

=> $r_{outer}=(\frac{3\times1.26}{g\times(e_{water})4\times3.14})^\frac{1}{3}$

Upon substitution of values of $e_{water}$ and g, we get: $r_{outer}=6.70cm$

Hence, the outer radius of the ball = 6.70cm

To learn more about Buoyant Force, refer to the link: brainly.com/question/17009786

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4 0

2 years ago

An alien spaceship streaks past Spanos football stadium along the direction of play at 0.8c. A football field is 120 yards long

Viktor [21]

Answer:

(i) Length of stadium, L₀ = 72 yards; width of stadium W = 55 yards (ii) 5 hours (iii) The people on earth. This is because they are not moving at relativistic speed relative to the game. (iv) The people on earth. This is because they are not moving at relativistic speed relative to the game.

Explanation:

(i) Given Length of stadium, L₀ = 120 yards, width of stadium, W₀ = 55 yards, speed of alien spaceship in direction of play, v = 0.8c

We use the equation for length contraction to determine the length of stadium L measured by the alien spaceship, since it is moving in the direction of game play and length contraction only occurs in the direction of motion. So, L = $\sqrt{1-\beta^{2} }$ L₀ where β = $\frac{v}{c}$ v= speed of alien spaceship and c= speed of light. So, β = $\frac{v}{c}$ = $\frac{0.8c}{c}$ = 0.8. Therefore, L = $\sqrt{1-0.8^{2} }$ L₀= $\sqrt{1-0.64}$ L₀= $\sqrt{0.36}$ L₀ = 0.6L₀ = 0.6 × 120 = 72 yards. So, the alien spaceship measures a length of 72 yards. The alien spaceship measures a width,W of 55 yards because there is no length contraction in the direction perpendicular to that of its motion. So, W = W₀ = 55 yards.

(ii) Since the game begins at 1:30 PM Pacific Standard Time (PST) and ends 4:30 PM Pacific Standard Time (PST), the proper time t₀, which is the duration of the event is 4:30 - 1:30 = 3 hours. The time measured by the alien spaceship t (since the time dilates)is given by t = t₀/ $\sqrt{1 - \beta ^{2} }$ . From (i) above β = 0.8 and t₀ = 3 hours. So, t = 3/ $\sqrt{1 - 0.8^{2} }$ = 3/0.6 = 5 hours.

(iii) The people in Spanos football stadium. This is because they are not moving at relativistic speed relative to the game. Since $\frac{v}{c}$ ≈ 0 then β ≈ 0 So, L = $\sqrt{1-\beta^{2} }$ L₀= So, L = $\sqrt{1-0}$ L₀thus L = L₀ for the spectators in spanos football field (iv) The people in Spanos football stadium. This is because they are not moving at relativistic speed relative to the game. Since $\frac{v}{c}$ ≈ 0 then β ≈ 0. So, t = t₀/ $\sqrt{1 - \beta ^{2} }$ = t₀/ $\sqrt{1-0}$ = t₀.Thus t = t₀ for the spectators in spanos football field.

8 0

3 years ago

Tether ball is a game children play in which a ball hangs from a rope attached to the top of a tall pole. The children hit the b

Murljashka [212]

Answer:

a_total = 14.022 m/s²

Explanation:

The total acceleration of a uniform circular motion is given by the following formula:

$a=\sqrt{a_c^2+a_T^2}$ (1)

ac: centripetal acceleration

aT: tangential acceleration

Then, you first calculate the centripetal acceleration by using the following formula:

$a_c=r\omega^2$

r: radius of the circular trajectory = 2.0m

w: final angular velocity of the ball = 7.0 rad/s

$a_c=(2.0m)(7.0rad/s)^2=14.0\frac{m}{s^2}$

Next, you calculate the tangential acceleration. aT is calculate by using:

$a_T=r\alpha$ (2)

α: angular acceleration

The angular acceleration is:

$\alpha=\frac{\omega_o-\omega}{t}$

wo: initial angular velocity = 13 rad/s

t: time = 15 s

Then, you use the expression for the angular acceleration in the equation (1) and solve for aT:

$a_T=r(\frac{\omega_o-\omega}{t})=(2.0m)(\frac{7.0rad/s-13.0rad/s}{15s})=-0.8\frac{m}{s^2}$

Finally, you replace the values of aT and ac in the equation (1), in order to calculate the total acceleration:

$a=\sqrt{(14.0m/s^2)^2+(-0.8m/^2)^2}=14.022\frac{m}{s^2}$

The total acceleration of the ball is 14.022 m/s²

5 0

3 years ago

Why is it important to know the center of gravity of an object?

Blizzard [7]

Explanation:

The direction of the force of gravity through the body is downward, towards the centre of the earth and through the COG. This line of gravity is important to understand and visualise when determining a person's ability to successfully maintain Balance.

7 0

3 years ago

The greatest speed with which an athlete can jump vertically is around 5 m/sec. Determine the speed at which Earth would move do

katrin2010 [14]

Answer:

Approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ if that athlete jumped up at $1.8\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81\; \rm m\cdot s^{-1}$ .)

Explanation:

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$ . That is: $p = m \cdot v$ .

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: $v(\text{athlete, before}) = 0$ and $v(\text{earth, before}) = 0$ . Therefore:

$\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned}$ and $p(\text{earth, before}) = 0$ .

Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.

Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:

$\begin{aligned}& p(\text{athlete, after}) + p(\text{earth, after}) \\ & =p(\text{athlete, before}) + p(\text{earth, before}) \\ &= 0\end{aligned}$ .

Therefore:

$p(\text{athelete, after}) = - p(\text{earth, after})$ .

$\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}$ .

Rewrite this equation to find an expression for $v(\text{earth, after})$ , the speed of the earth after the jump:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}$ .

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is $g = 9.81\; \rm N \cdot kg^{-1}$ .

$\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}$ .

Calculate $v(\text{earth, after})$ using $m(\text{earth})$ and $v(\text{athlete, after})$ values from the question:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}$ .

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ .

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3 years ago