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mylen [45]
4 years ago
15

An individual experiences a deep cut that severs the radial artery near the elbow. Ignoring air resistance, approximately how hi

gh will the blood spurt? (Hints: the specific gravity of blood is 1.050 g cm-3 and the specific gravity of mercury is 13.6 g cm-3 )
Physics
1 answer:
tatiyna4 years ago
4 0

Hey! How are you? My name is Maria, 19 years old. Yesterday broke up with a guy, looking for casual sex.

Write me here and I will give you my phone number - *pofsex.com*

My nickname - Lovely

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What happens when a star comes to an end?
nlexa [21]

Answer:

It depends on the size of the star but I think it's B

Explanation:

If the star is massive, it will eventually explode (supernova) and if it is a star with a high mass, the core of it will form a neutron star and if it is very massive the core will turn into a blackhole

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Which of the following is an example of the transformation of gravitational potential energy into motion energy (kinetic energy
nikklg [1K]

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A drop of water falling from a faucet into a sink

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In a coal-fired power plant, after the steam has been used to turn the turbine to power the generator, it goes through a condens
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give me theze points

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3 years ago
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The oil level in a tank is 2 m above the ground. The tank cover is air tight and the air pressure above the oil surface is 150 k
Andrew [12]

Answer:

a) 24.692 m/s

b) 19.4 m

Explanation:

To calculate the velocity at the nozzle outflow (V2) we use the Bernoulli equation:

\frac{P1}{pg}+\frac{V1^2}{2g}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}+Z2

We know that the velocity above the oil surface (V1) and the pressure at the nozzle outflow (P2) are negligible, the height in the exit is zero (Z2) then:

\frac{P1}{pg}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}

a) The velocity (V2) is:

\frac{P1}{pg}+Z1=\frac{V2^2}{2g}

(\frac{P1}{pg}+Z1)(2g)=V2^2

V2=[(\frac{P1}{pg}+Z1)(2g)]^{1/2}

Substituting the known values we can get the velocity at the out:

Atmospheric pressure= 101000 Pa

Oil density= 0.88x(Water density)=0.88(1000kg/m3)=880kg/m3

V2=[(\frac{150000Pa+101000 Pa}{(880 kg/m3)(9.81m/s)}+2m)(2(9.81m/s2))]^{1/2}

V2=24.692 m/s

b) To calculate the height we have to apply the Bernoulli equation between the outflow and the maximum height (Z3), so:

\frac{P2}{pg}+\frac{V2^2}{2g}+Z2=\frac{P3}{pg}+\frac{V3^2}{2g}+Z3

We know that the velocity above the stream (V3) and the pressure at the nozzle outflow (P2) are negligible, the pressure at the top of the stream (P3) is the atmospheric pressure, then:

\frac{V2^2}{2g}=\frac{P3}{pg}+Z3

Z3=\frac{V2^2}{2g}-\frac{P3}{pg}

Substituting the known values, the height (Z3) is:

Z3=\frac{(24.692 m/s)^2}{2(9.81 m/s2)}-\frac{101000 Pa}{(9.81 m/s)(880 kg/m3)}

Z3=Maximum Height=19.376=19.4 m

3 0
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How might acid precipitation affect local economies?
ser-zykov [4K]
For example if they produce limestone or quarry it or local buildings made out of limestone rock could erode under acid rain
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