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FinnZ [79.3K]
3 years ago
14

The volume flow rate in an artery supplying the brain is 4.00 10-6 m3/s. (a) If the radius of the artery is 4.50 10-3 m, determi

ne the average blood speed. m/s (b) Find the average blood speed at a constriction in the artery if the constriction reduces the radius by a factor of 4. Assume that the volume flow rate is the same as that in part (a).
Physics
1 answer:
mr_godi [17]3 years ago
8 0

Answer:

(a) 0.063 m/s

(b) 1.01 m/s

Explanation:

rate of volume flow, V = 4 x 10^-6 m^3/s

(a) radius, r = 4.5 x 10^-3 m

Let the speed of blood is v.

So, V = A x v

where A be the area of crossection of  artery

4 x 10^-6 = 3.14 x 4.5 x 10^-3 x 4.5 x 10^-3 x v

v = 0.063 m/s

Thus, the speed of flow of blood is 0.063 m/s .

(b) Now r' = r / 4 = 4.5 /4 x 10^-3 m = 1.125 x 10^-3 m

Let the speed is v'.

So, V = A' x v'

4 x 10^-6 = 3.14 x 1.125 x 10^-3 x 1.125 x 10^-3 x v'

v' = 1.01 m/s

Thus, the speed of flow of blood is 1.01 m/s .

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If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock com
Lelu [443]

Answer:

<em> B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.</em>

<em></em>

Explanation:

This is the complete question

A person will feel a shock when a current of greater than approximately 100μ A flows between his index finger and thumb. If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock compare in each scenario?

A) The voltage on dry skin needs to be 200 times smaller than the voltage on wet skin.

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

C) The voltage on dry skin is the same as the voltage on wet skin.

D) The voltage on dry skin needs to be 40,000 times larger than the voltage on wet skin.

Ohm's law states that electric current is proportional to voltage and inversely proportional to resistance.

the equation is written as

V = IR

Where V is the voltage

I is the current

R is the resistance

for this case, the current I is 100μ A = 100 x 10^16 A

resistance of wet skin = R

resistance of dry skin = 200R

for the wet skin, voltage will be

V = IR = 100*10^{-6} R

for dry skin, voltage will be

V = IR = 100*10^{-6}*200R = 0.02R

Comparing both voltages

0.02R ÷  100*10^{-6} R  = 200

<em>this means that the voltage on the wet skin should be 200 times lesser than the voltage on the dry skin or the voltage on the dry skin should be 200 times more than the voltage on the wet skin.</em>

4 0
3 years ago
If the Earth were compressed in such a way that its mass remained the same, but the distance around the equator were just one-ha
vovikov84 [41]
If the distance around the equator is reduced by half, then the radius is also reduced by half.

Since the acceleration due to gravity is proportional to 1/(radius²),
the acceleration changes by a factor of 1/(1/2)² = 1/(1/4) = <em>4 </em>.

The acceleration due to gravity ... and also the weight of everything on Earth ...
becomes <em>4 times what it is now</em>.
6 0
3 years ago
Read 2 more answers
You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of
Sunny_sXe [5.5K]

Answer:

P = 1.090\,N

Explanation:

The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

\Sigma F = P - W + F_{D}

\Sigma F = P - \rho \cdot V \cdot g + c\cdot v = 0

Now, the exerted force is:

P = \rho \cdot V \cdot g - c\cdot v

The volume of a sphere is:

V = \frac{4\cdot \pi}{3}\cdot R^{3}

V = \frac{4\cdot \pi}{3}\cdot (0.014\,m)^{3}

V = 1.149\times 10^{-5}\,m^{3}

Lastly, the force is calculated:

P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )

P = 1.090\,N

5 0
3 years ago
What happens if you reverse the galaxy motion and go back in time?
bulgar [2K]
It’s called mirror universe
7 0
3 years ago
Please help me with this (with explanation)
Sergeeva-Olga [200]

Suppose the cyclist travels for a total time of <em>t</em> hours.

For 20 min = 1/3 hr, the cyclist does not move.

Over the remaining (<em>t</em> - 1/3) hr, the cyclist is moving at a constant speed of 22.0 km/hr, so that the cyclist would travel a distance of

<em>x</em> = (22.0 km/hr) • ((<em>t</em> - 1/3) hr) ≈ (22.0 km/hr) <em>t</em> - 7.33 km

If the cyclist's average speed over the total time <em>t</em> was 17.5 km/hr, then by the definition of average speed,

17.5 km/hr = <em>x</em> / <em>t</em>

Replace <em>x</em> with the distance expression from earlier:

17.5 km/hr = ((22.0 km/hr) <em>t</em> - 7.33 km) / <em>t</em>

Solve for <em>t</em> :

17.5 km/hr = 22.0 km/hr - (7.33 km) / <em>t</em>

(7.33 km) / <em>t</em> = 4.5 km/hr

<em>t</em> = (7.33 km) / (4.5 km/hr)

<em>t</em> ≈ 1.62963 hr

Then the distance the cyclist traveled must have been

<em>x</em> ≈ (22.0 km/hr) (1.62963 hr) - 7.33 km ≈ 28.5 km

and so the answer is A.

Alternatively, as soon as you arrive at

17.5 km/hr = <em>x</em> / <em>t</em>

you can instead solve for <em>t</em> in terms of <em>x</em>, then plug that into the distance equation.

<em>t</em> = <em>x</em> / (17.5 km/hr)

then

<em>x</em> ≈ (22.0 km/hr) (<em>x</em> / (17.5 km/hr)) - 7.33 km

<em>x</em> ≈ 1.25714 <em>x</em> - 7.33 km

0.25714<em>x</em> ≈ 7.33 km

<em>x</em> = (7.33 km) / 0.25714 ≈ 28.5 km

6 0
2 years ago
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