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vaieri [72.5K]
3 years ago
12

"A 78-year-old woman has a mean arterial pressure of 120 mm Hg and a heart rate of 60 beats/min. She has a stroke volume of 50 m

L, cardiac output of 3000 mL/min, and a right atrial pressure of 0 mm Hg. What is the total peripheral resistance (in mm Hg/mL/min) in this woman
Physics
1 answer:
aivan3 [116]3 years ago
8 0

Answer:

0.04 mm Hg / mL / min .

Explanation:

Arterial pressure = 120 mm Hg

right atrial pressure = 0 mm Hg

Drop in pressure due to peripheral resistance = 120 mm Hg

volume of cardiac output per minute = 3000 mL/min

total  peripheral resistance

= 120 / 3000 mm Hg / mL / min

= 0.04 mm Hg / mL / min .

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swimmers at a water park have a choice of two frictionless water slides as shown in the figure. although both slides drop over t
LuckyWell [14K]
If swimmers had a choice of the water slides shown in this figure,
they would all go home dry, since there is no figure.  I'll have to try to
answer this question based on only the words in the text, augmented
only by my training, education, life experience, and human logic.

-- Both slides are frictionless.  So no energy is lost as a swimsuit
scrapes along the track, and the swimmer's kinetic energy at the
bottom is equal to the potential energy he had at the top.

-- Both slides start from the same height.  So the same swimmer
has the same potential energy at the top of either one, and therefore
the same kinetic energy at the bottom of either one.

-- So the difference in the speeds of two different swimmers
on the slides depends only on the difference in the swimmers'
mass, and is not influenced by the shape or length of the slides
(as long as the slides remain frictionless).

If both swimmers have the same mass, then  v₁ = v₂ .
4 0
4 years ago
A small glass bead has been charged to 8.0 nc. what is the magnitude of the electric field 2.0 cm from the center of the bead?
astraxan [27]
<span>Charge of the glass bead Q = 8.0 x 10^-9 C Distance d = 2.0 cm = 0.02 m Coulombs constant K = 8.99 x 10^9 Nm^2/C^2 Electric Field E = k x Q / d^2 = 8.99 x 10^9 x 8.0 x 10^-9 / (0.02)^2 E = 71.92 / 0.0004 = 17.98 x 10^4 The electric field is 1.8 x 10^5 N/C</span>
7 0
3 years ago
Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax
kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its x-component is

F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its x-component is

F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N

So, the x-component of the resultant force is

F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its y-component is

F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its y-component is

F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N

So, the y-component of the resultant force is

F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

F=\sqrt{F_x^2+F_y^2}

Where in this problem, we have:

F_x=-7.14 N is the x-component

F_y=2.70 N is the y-component

And substituting, we find:

F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N

6 0
3 years ago
Read 2 more answers
Fighter jet starting from airbase A flies 300 km east , then 350 km at 30° west of north and then 150km north to arrive finally
Tems11 [23]
A jet fighter flies from the airbase A 300 km East to the point M. Then 350 km at 30° West of North.
It means : at 60° North of West. So the distance from the final point to the line AM is :
350 · cos 60° = 350 · 0.866 = 303.1 km
Let`s assume that there is a line N on AM.
AN = 125 km and NM = 175 km.
And finally jet fighter flies 150 km North to arrive at airbase B.
NB = 303.1 + 150 = 453.1 km
Then we can use the Pythagorean theorem.
d ( AB ) = √(453.1² + 125²) = √(205,299.61 + 15,625) = 470 km
Also foe a direction: cos α = 125 / 470 = 0.266
α = cos^(-1) 0.266 = 74.6°
90° - 74.6° = 15.4°
Answer: The distance between the airbase A and B is 470 km.
Direction is : 15.4° East from the North. 
4 0
3 years ago
Are these two correct ?
gulaghasi [49]

Answer:

7. Your answer is correct dear, just add the unit

8. answer is 1.17m/s²

Explanation:

queation 7.

m = 3kg, F = 9N, a ?

F = ma

a = F/m = 9/3 = 3m/s²

Use the same approach for question 8

5 0
3 years ago
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