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sergeinik [125]
2 years ago
12

When you hold your hands at your sides, you may have noticed that the veins sometimes bulge--the height difference between your

heart and your hands produces increased pressure in the veins. The same thing happens in the arteries. If the average arterial pressure at your heart is a typical 100 mmHg, what is the average arterial pressure in your hands when they are held at your side? Assume your hands are 60 cm below your heart.
Physics
1 answer:
hodyreva [135]2 years ago
7 0

To solve this problem it is necessary to use the concepts related to pressure and pressure, absolute and atmospheric.

Average arterial pressure in the hands,

P = 100mmHg+ h*\rho_{blood}g

Where,

P = Pressure

h = height (at this case the length of the arm)

Replacing with our values

P = 100mm(Hg)+(600mm)(\frac{\rho_{blood}}{\rho_{mercury}})(Hg)

P = 100mmHg+600*\frac{1060}{13600}mmHg

P = 100mmHg+46.765mmHg

P = 146.765mmHg

Where,

\rho_{blood} = 1060Kg/m^3

\rho_{mercury}=13600Kg/m^3

Therefore the pressure is 146.765mmHg

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Sharks are generally negatively buoyant; the upward buoyant force is less than the weight force. This is one reason sharks tend
Tresset [83]

Answer:

8.67807 N

34.7123 N

Explanation:

m = Mass of shark = 92 kg

\rho_{se} = Density of seawater = 1030 kg/m³

\rho_{f} = Density of freshwater = 1000 kg/m³

\rho_{sh} = Density of shark = 1040 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Net force on the fin is (seawater)

F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N

The lift force required in seawater is 8.67807 N

Net force on the fin is (freshwater)

F_n=mg-V_s\rho_{f}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{f}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1000\times 9.81\\\Rightarrow F_n=34.7123\ N

The lift force required in a river is 34.7123 N

6 0
2 years ago
A 10kg rocket is traveling at 80 m/s when the booster engine applies a constant forward force of 60 N for 3.0 seconds. What impu
Lina20 [59]

Answer:

Impulse = 90

Resulting Velocity = 89

Explanation:

Use F * change in time = m * change in velocity.

For the first part of the question, the left side of the equation is the impulse. Plug it in.

60 * (3.0 - 0) = 90.

For the second half. we use all parts of the equation. I'm gonna use vf for the final velocity.

60 * (3.0 - 0) = 10 * (vf - 80). Simplify.

90 = 10vf - 800. Simplify again.

890 = 10vf. Divide to simplify and get the answer.

The resulting velocity is 89.

4 0
2 years ago
How do air mass conditions ahead of the squall line support the development of new cell?
IRISSAK [1]
<span>Storm cells in a squall line typically move from the southwest to the northeast, and as the mature cells in the northeast begin to die off, new ones are formed at the opposite end to advance the line. The air in the southwest corner has strong vertical updrafts that allow new cells to grow and develop into thunderstorms.</span>
7 0
2 years ago
A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in.2 The
WINSTONCH [101]

Answer:

31.67 in

Explanation:

Given:

Diameter of the pipe, D = 3ft = 36 in

cross-sectional area of the steel = 0.2 in²

Note: Refer to the figure attached

From the free body diagram represented in the figure, we have

ΣFx = 0

or

pressure × projected area = 2 × Force in steel

Now, the projected area = spacing (s) × diameter of the wood pipe

force in steel = stress in steel (σ) × cross-sectional area of the steel

on substituting the values we get

4 ksi × (s × 36 in) = 2 × σ × 0.2 in²

also, allowable hoop stress = 11.4 ksi

thus,

σ = 11.4 ksi = 11.4 × 10³ psi

therefore, we have

4 psi × (s × 36 in) = 2 × 11.4 × 10³ psi × 0.2 in²

thus,

s = 31.67 in

hence the maximum spacing is 31.67 in

3 0
2 years ago
read the excerpt below and answer the question. "no roving foot shall crush thee here, no busy hand provoke a tear." what type o
Aneli [31]
The type of figurative language that Freneau employ in these lines from "The Wild Honeysuckle" is personification. The correct answer would be option D. Why is it personification? Personification is a figure of speech that uses human attributes to something that is not living. In this line, the author attributes the words "roving" and "busy" to foot and hand, respectively. 
3 0
2 years ago
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