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agasfer [191]
4 years ago
11

The aorta is approximately 25 mm in diameter. The mean pressure there is about 100 mmHg and the blood flows through the aorta at

approximately 60 cm/s. Suppose that at a certain point a portion of the aorta is blocked so that the cross-sectional area is reduced to 3/4 of its original area. The density of blood is 1060 kg/m3.
(a) How fast is the blood moving just as it enters the blocked portion of the aorta? (in cm/s)

(b) What is the gauge pressure (in mmHg) of the blood just as it has entered the blocked portion of the aorta? (in mmHg)
Physics
1 answer:
Ksenya-84 [330]4 years ago
8 0

Answer:

Explanation:

25 mm diameter

r₁  = 12.5 x 10⁻³ m radius.

cross sectional area =  a₁

Pressure P₁  = 100 x 10⁻³ x 13.6 x 9.8 Pa

a )

velocity of blood v₁ = .6 m /s

Cross sectional area at blockade = 3/4 a₁

Velocity at blockade area = v₂

As liquid is in-compressible

a₁v₁ = a₂v₂

a₁ x .6 m /s  = 3/4 a₁ v₂

v₂ = .8m/s

b )

Applying Bernauli's theorem formula

P₁ + 1/2 ρv₁² =  P₂ +  1/2 ρv₂²

100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ +  1/2x 1060 x .8²

13328 +190.8 = P₂ + 339.2

P₂ = 13179.6 Pa

= 13179 / 13.6 x 10³ x 9.8 m of Hg

P₂ =  .09888 m of Hg

98.88 mm of Hg

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klio [65]

Answer:

The unit that is a radical among the options is;

a. NH₄⁺

Explanation:

A radical is an atom, group of atoms molecule or ion that contains an electron that is unpaired. The presence of the unpaired electrons make radical very reactive as such most radicals enter into reactions with other species or with themselves to form new compounds making them not last long.

Radicals are also known as free radicals. Examples of radicals includes; Ammonium ion, NH₄⁺

Nitrate ion NO₃⁻ (From a salt)

Hydroxide ion OH⁻ (From a base)

Sulphate ion SO₄²⁺ (From an acid)

All of which are formed in aqueous solution of acids, salts and bases.

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4 years ago
3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between
Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

v^2 -u^2 = 2ad (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

F=ma=-\mu mg

where

m = 1000 kg is the mass of the car

\mu = 0.80 is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

a=-\mu g

And we can substitute it into eq.(1) to find d:

v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m

7 0
4 years ago
WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of
nekit [7.7K]

Answer:

82.1 km

Explanation:

We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:

A_x = (72) sin 30^{\circ} =36.0 km\\A_y = (72) cos 30^{\circ} = 62.4 km

Vector B is 48 km south, so:

B_x = 0\\B_y = -48

Finally, vector C:

C_x = -(100) cos 30^{\circ} =-86.6 km\\C_y = (100) sin 30^{\circ} = 50.0 km

Now we add the components along each direction:

R_x = A_x + B_x + C_x = 36.0 + 0 +(-86.6)=-50.6 km\\R_y = A_y+B_y+C_y = 62.4+(-48)+50.0=64.6 km

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R=\sqrt{R_x^2+R_y^2}=\sqrt{(-50.6)^2+(64.6)^2}=82.1 km

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3 years ago
Which is better, ramen, or spaghetti
Sergio039 [100]

Answer:

ramen

Explanation:

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3 years ago
Read 2 more answers
Two point charges of equal magnitude are 8.0 cm apart. At the midpoint of the line connecting them, their combined electric fiel
bagirrra123 [75]

Answer:

r = 8/2 = 4cm = 0.04m

k = 9×10^9

Enet = 51 N/C

Enet = E1 + E2

since E1 = E2

E1 = Enet/2 = 51/2

E/2 = kq/r²

q = Er²/2k

q = (51 × 0.04²)/(2×9×10^9)

q = 4.5×10^-12 C

q1 = q2 = 4.5 pC

Explanation:

The electric field is a region around a

charge in which it exerts electrostatic force

on another charges. While the strength of

electric field at any point in space is called

electric field intensity. It is a vector

quantity. Its unit is NC¯¹.

According to coulomb’s law ,if a unit

positive charge q (call it a test charge) is

brought near a charge q (call a field

charge) placed in space,the charge q will

experience a force. The value of this force

depends upon the distance between the

two charges. If the charge q is moved

away from q ,this force would decrease till

at a certain distance the force would be

practically reduced to zero. The charge q

is then out of the influence of charge q.

The region of space surrounding the charge

q in which it exerts a force on the charge

q is known as E.F of the charge

q. Mathematically it is expressed as:

E =F/q

The direction of the vector E is the same

as the direction of F,because q is a

positive scalar. Dimensionally,the E.F is

force per unit charge,and its SI unit is the

newton/coulomb (N/C).

7 0
3 years ago
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