Ask question

Ask question

All categories

3 years ago

7

If the radius of a blood vessel drops to 84.0% of its original radius because of the buildup of plaque, and the body responds by

increasing the pressure difference across the blood vessel by 10.0%, what will have happened to the flow rate? The flow rate will have changed to ...... % of its original value.

1 answer:

erma4kov [3.2K]3 years ago

4 0

To develop this problem it is necessary to apply the equations concerning Bernoulli's law of conservation of flow.

From Bernoulli it is possible to express the change in pressure as

$\Delta P = \frac{1}{2}\rho (v_1^2-v_2^2)+ \rho g (h_1h_2)$

Where,

$v_i =$ Velocity

$\rho =$ Density

g = Gravitational acceleration

h = Height

From the given values the change of flow is given as

$R = r^4P$

Therefore between the two states we have to

$\frac{R_2}{R_1} = \frac{r_2^4 P_2}{r_1^4 P_1} *100\%$

$\frac{R_2}{R_1} = \frac{84^4 (110)}{100^4*(100)} *100\%$

$\frac{R_2}{R_1} = 54.77\%$

The flow rate will have changed to 54.77 % of its original value.

You might be interested in

Explain relative velocity.

xxTIMURxx [149]

Answer:

The relative velocity of an object A with respect to another object B.

Explanation:

The relative velocity of an object A with respect to another object B is the velocity that object A would appear to have to an observer situated on object B moving along with it.

4 0

3 years ago

ou purchase a rectangular piece of metal that has dimen- sions 5.0 * 15.0 * 30.0 mm and mass 0.0158 kg. The seller tells you tha

Natalija [7]

Answer: 7022.2kg/m³, yes, I was cheated

Explanation:

Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;

Density = Mass/Volume

Note that the unit of both mass and volume must be standard unit.

Given mass = 0.0158kg

Dimension of the metal = 5mm×15mm×30mm

Note that 1mm = 0.001m

The volume of the metal will be

0.005×0.015×0.03

= 0.00000225m³

Density = 0.0158/0.00000225

Average density of the metal = 7022.2kg/m³

Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.

4 0

3 years ago

PLEASE HELPPPPPPPPPP

olga55 [171]

Answer:

Peer Review

Explanation:

When scientist check other scientists the process is dubbed "peer review".

8 0

3 years ago

During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±

Alla [95]

Answer:

P = 1 (14,045 ± 0.03 ) k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

Δ (Pₓ / Py) =?

Let's start by finding the momentum of each vehicle

car X

Pₓ = m vₓ

Pₓ = 2.34 2.5

Pₓ = 5.85 kg m

car Y

Py = 2,561 3.2

Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

ΔPₓ = m Δv + v Δm

ΔPₓ = 2.34 0.01 + 2.561 0.01

ΔPₓ = 0.05 kg m

Δ $P_{y}$ = m Δv + v Δm

ΔP_{y} = 2,561 0.01+ 3.2 0.001

ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

P = Pₓ / $P_{y}$

ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

ΔP = 0.006 + 0.0026

ΔP = 0.009 kg m

The result is

P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s

7 0

3 years ago

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

$V_{Ax}=V_{A}cos\theta$

$V_{Ax}=(21m/s)cos(-14^{o})$

$V_{Ax}=20.38 m/s$

$V_{Ay}=V_{A}sin\theta$

$V_{Ay}=(21m/s)sin(-14^{o})$

$V_{Ay}=-5.08 m/s$

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

$x_{A}=V_{Ax}t$

$x_{A}=(20.38m/s)(0.317s)$

$x_{A}=6.46m$

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

$V_{Bx}=20.88 m/s$

$V_{By}=-2.195 m/s$

$y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}$

$0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}$

$-4.9t^{2}-2.195t+2.1=0$

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}$

t= -0.9159s or t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

$x_{B}=V_{Bx}t$

$x_{B}=(20.88m/s)(0.468s)$

$x_{B}=9.77m$

So once we got the two distances we can now find the difference between them:

$x_{B}-x_{A}=9.77m-6.46m=3.31m$

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0

3 years ago