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AysviL [449]
3 years ago
8

Read the following characteristics of stars and choose the one that is correct. plz help quickly

Physics
1 answer:
aleksklad [387]3 years ago
4 0
The answer to this question will D
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Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of
Sever21 [200]

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) <em>High and low pressures in kilopascals</em>:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}

p_{high} = 15.999\,kPa

p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}

p_{low} = 10.666\,kPa

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) <em>High and low pressures in pounds per square inch</em>:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}

p_{high} = 2.320\,psi

p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}

p_{low} = 1.547\,psi

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) <em>High and low pressures in meter water column in meters water column</em>:

We can calculate the equivalent water column of a mercury column by the following relation:

\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}

h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg} (Eq. 1)

Where:

\rho_{w}, \rho_{Hg} - Densities of water and mercury, measured in kilograms per cubic meter.

h_{w}, h_{Hg} - Heights of water and mercury columns, measured in meters.

If we know that \rho_{w} = 1000\,\frac{kg}{m^{3}}, \rho_{Hg} = 13600\,\frac{kg}{m^{3}}, h_{Hg, high} = 0.120\,m and h_{Hg, low} = 0.080\,m, then we get that:

h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m

h_{w, high} = 1.632\,m

h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m

h_{w, low} = 1.088\,m

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

4 0
3 years ago
Which of the following types of technologies has best helped scientists to study very high-energy objects in outer space, such a
erica [24]

Very high-energy objects and events spit out very high-energy photons, so the instrument you need in order to detect them is the       X-ray telescope. <em>(C)  </em>

Inconveniently, X-ray telescopes only work when they're up in orbit, because X-rays get seriously soaked up in Earth's atmosphere, and most of them never make it down to the surface ... (lucky for us !) .

3 0
3 years ago
Read 2 more answers
Pls help on this one?
sattari [20]
The answer is point C
4 0
2 years ago
A 1-kilogram ball has a kinetic energy of 50 joules the speed of the ball is
sveta [45]

Answer:

10 m/s

Explanation:

3 0
2 years ago
A skater with a mass of 72 kg is traveling east at 5.8 m/s when he collides with another skater of mass 45 kg heading 60° south
Akimi4 [234]

The final velocity is 5.87 m/s

<u>Explanation:</u>

Given-

mass, m_{1} = 72 kg

speed, v_{1} = 5.8 m/s

Mass_{2},m_{2}  = 45 kg

speed_{2},v_{2}  = 12 m/s

Θ = 60°

Final velocity, v = ?

Applying the conservation of momentum:

m_{1} X v_{1} + m_{2} X v_{2} = (m_{1} +m_{2} ) v

72 X 5.8 + 45 X 12 X cos 60° = (72 + 45) v

v = 417.6 + 540 X \frac{0.5}{117}

v = 417.6 + \frac{270}{117}

v = 5.87 m/s

The final velocity is 5.87 m/s

8 0
3 years ago
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